
Class 

Book 

Copyright W. 



S6 



COPYRIGHT DEPOSIT. 



In Press 
BY THE SAME AUTHOR 



A MANUAL 

OF THE 

METRIC SYSTEM 

Showing by comparisons the sim- 
plicity of the construction of 
the system and the ad- 
vantages to be obtained 
by its use 



A HAND BOOK 

OF 

General Instruction 

FOR 

MECHANICS 



CONTAINING 

USEFUL RULES AND MEMORANDA 

FOR PRACTICAL MEN 



BY 

FRANKLIN E. SMITH 



NEW YORK 

D. VAN NOSTRAND COMPANY 

23 MURRAY AND 27 WARREN STS. 
I907 



Copyright, 1906 
By D. Van Nostrand Company 

COPYKI«aT 1907, 

■T 

B. Van Nost&and Company , 



?U3RARYofCONGRiIs| 

Two Cooles Received 

AUG 2o I90f 

Cooyrifht Entry 

CLASS 4 XXc./No, 

COPY B. 



T4 



•Sco 



T/ie Plimpton Press Norwood Mass. U.S.A. 



PREFACE 

The primary object of the author in writing this book 
is to give the mechanic, who has not had educational 
advantages, a text-book explaining established rules for 
calculating in a clear, simple, and concise way, making 
him familiar with the various technical terms and their 
meaning, and to be in general such a course of instruction 
as to impart, in a simple manner, the required knowledge 
to enable him to read understandingly more advanced 
scientific works. 

In writing the book, no attempt has been made to 
limit the language to the rigid brevity of the average 
text-book on mathematics and physics, a very free style 
being used which will be more acceptable to the general 
reader. 

Franklin E. Smith. 



CONTENTS 

PART I 

Arithmetic 

i 

PAGE 

Numeration » 4 

Notation 9 

Addition 10 

Subtraction 16 

Multiplication 22 

Division '. 40 

Tables of Weights and Measures Used by Mechanics .- . . 63 

Reduction 67 

Fractions 72 

Addition of Fractions . 85 

Arithmetical Signs 85 

Subtraction of Vulgar Fractions 94 

Multiplication of Vulgar Fractions 97 

Division of Vulgar Fractions 99 

Decimal Fractions 103 

Reduction of Decimal Fractions 108 

Addition of Decimals 109 

Subtraction of Decimals 110 

Multiplication of Decimals Ill 

Division of Decimals 113 

Proportion 119 

Compound Proportion 126 

Interest 131 

Involution 137 

Evolution .138 

Cube Root 141 



vi CONTENTS 



PART II 

Arithmetical Signs and Characters and Explanation of 
Solving Formula 

page 

Arithmetical Signs 149 

Examples on the Use of the Addition Signs 151 

Examples Performed where Addition and Minus Signs are Used 152 

Examples Showing how Brackets are Used 154 

Examples on Proper Use of Multiplication Sign .... 156 

Examples on Proper Use of Division Sign 158 

Examples on Signs Representing the Powers of Numbers . 161 

Examples on Signs that Represent the Roots of Numbers . 162 

Law of Signs in Multiplication 164 

Law of Signs in Division 166 

Formula 167 



PART III 

Mensuration 

Table of Decimal Ecfuivalents 177 

Explanation of Terms 177 

The Circle 182 

The Ellipse 184 

The Triangle 187 

The Rhomboid 192 

The Trapezium 196 

The Sphere 199 

Volume Measure and Contents of Solids 200 

Rectangular Solids 200 

The Cylinder 201 

The Pyramid 202 

The Cone 204 

The Frustum 205 



CONTENTS vii 

PART IV 

Weight 

> Specific Gravity 

How the Dimensions, Measurements, and Weight of 

Different Shaped Vessels is Found 

and 

How the Weight of Different Parts is Found 

page 

Specific Gravity 211 

Table of Specific Gravities of Liquids and Solids . . . .214 

Circular Measure 219 

Square Measurements 224 

Circular Areas 229 

Measurements and Weights of Tanks 234 

Measurements of Circular Tanks ......... 238 

How to Calculate the Weight of Different Materials and the 
Weight of the Different Parts of a Machine or Structure of 

any Kind 242 

Table of Weights of Various Metals .251 

PART V 
The Primary or Simple Machines 

The Lever 256 

The Compound Lever 268 

The Pulley 269 

The Strength of a Rope 275 

The Wheel and Axle 276 

The Inclined Plane . 284 

The Wedge 288 

The Screw 290 

PART VI 

Strength of Materials and Questions Relating to Stress 

Table Showing Ultimate and Elastic Strength of Materials . 308 

Table Showing Safe Working Stress 309 

Table of Factors of Safety 310 

Tensial Stress 311 

Compressive Stress 313 

Shearing Stress 315 



PART I 

ARITHMETIC 



ARITHMETIC 

Arithmetic is the science of numbers and has been 
known to man in various forms for all time. It is some- 
times called the ''science of calculation/' and when so 
termed the early period of the science is referred to when 
(calculi) pebbles were used by the savages to make the 
process of counting, computing, and estimating easy. 
In Arithmetic there are ten characters called figures: 

1234567 89 
Named: one two three four five six seven eight nine nought 

And each figure, with the exception of 0, represents a 
specific collection of single things called units. 

By unit is meant a single thing, as one cent, one dollar, 
one ounce, one pound, etc. 

Figure 1 represents one unit. 

Figure 2 represents a collection of two units. 

Figure 3 represents a collection of three units. 

The figure stands for nothing, and when alone ex- 
presses no units. 

A unit or collection of units is called a number, and 
a figure is a character used to represent a number; that 
is, figure 1 may be termed number 1, and figure 9 may 
be termed number 9. 

A number, then, is a unit or a collection of units 

Number 1 is called a unit. Number 9 is a collection 
of nine units. Number 2 is a collection of two units. 
Number 3 is a collection of three units, etc. 

3 



4 A HAND BOOK FOR MECHANICS 

By means of the figures 1, 2, 3, 4, etc., any known 
number may be represented. 

There are also six rules according to which arithmetical 
operations are performed, called 

Numeration. 

Notation. 

Addition. 

Subtraction. 

Multiplication. 

Division. 

The last four of these rules are called the fundamental 
rules of Arithmetic, and upon them depend the solution 
of all arithmetical problems. 

NUMERATION 

Numeration is the art of reading numbers which are 
written in figures. The art of naming the value of 
figures. 

Example. Naming the value of 9376267, for instance. 

Figures have two values called Simple value and Place 
value. 

The simple value of a figure is the value it has when 
standing alone, or on the right hand of a number of 
figures. 

In both cases the figure denotes the number of units 
it represents. 

For example, 2 standing alone, or when standing on 
the right hand of a number of figures, thus, 342, denotes 
in both two units or ones. 

The place value of a figure is the value it denotes 
when standing with other figures. And the place value 
of a figure depends upon its relative position to another 
figure. 



NUMERATION 5 

For example, each figure of the following number, 
1364, has a place value. And that value depends upon 
the number of places it is toward the left hand from the 
right-hand figure 4. 

Each removal of one place to the left increases the 
value of a figure ten times. 

For example, 4 standing at the left hand of 4, thus, 
44, expresses ten times the value it does when standing 
alone, or in the right-hand place. 

Standing alone or on the right-hand place it denotes 
four units only, but being removed one place to the left, 
as above, increases its value ten times, and it then de- 
notes ten times four, which is forty, and the four on the 
right of it denotes four units. And the two figures 
then together express forty-four (forty-four units or ones). 

Again, by removing the figure 4 one place more to 
the left, making it then three places from the right 
thus, 444, its value is again increased tenfold. 

The first removal to the left increases it from 4 to 40, 
and the next or third removal increases its value ten 
times more, making four hundred, and the three figures 
together express four hundred forty-four. 

Each place has a name, as follows, reading from right 
to left: 



03 

73 



PQ d 

\2 ° 

O S 

<D «+h 03 

i; o fl 

^ m Q 



W H PQ 



a 






c3 












o 






03 


«j 










1 


co 

a 

.2 




3 
O 
Jd 
H 


73 

i 

co 
O 

r-j 

H 










o 

CO 

73 


1 




o 

co 
73 


03 

73 

d 

c3 


CO 

73 






CP 


M-c 


03 


03 


«*-< 


cu 






i- 


o 


£ 


Sm 


o 


(-1 






73 




O 


73 




3 

O 

H 


73 




% 




co 

05 


9 


d 
3 


CO 


d 


CO 

d 

CD 

H 


'8 
P 



0, 0, 0, 0, 



6 A HAND BOOK FOR MECHANICS 

Often the places occupied by figures are called orders, 
thus, the right-hand or unit figure is called a figure of 
the first order, or the order of units. 

A figure in the second place is a figure of the second 
order, or the order of tens. 

In the third place, of the order of hundreds, etc. 

Each place name expresses a specific value or number. 
As, unit means the first whole number and expresses one. 

Ten means ten times one, or a value equal to ten ones. 

Hundred means ten times ten, or a value equal to ten 
times ten, single things, individuals or units. 

Thousand means ten hundred single things or units. 

Ten thousand means ten thousand single things or 
units. 

Hundred thousand means hundred thousand single 
things or units. 

Million means a thousand thousand single things or 
units. 

Ten million means ten thousand thousand single things 
or units. 

Hundred million means ten thousand thousand single 
things or units. 

Billion means one thousand million (1,000,000,000). 

Each figure of a number expresses so many units of 
that order to which it belongs. 

For example, one figure standing alone is called a 
unit, but when two of them stand together the left-hand 
one expresses so many units of the second order, which 
is tens, but the right-hand one is still called a unit; 
thus, 79 is a collection of nine units and seven sets of 
ten units each, or of nine units and seventy units, or of 
79 units, and is read seventy-nine. 

Where three figures stand together, thus, 346, the 
third one from the unit figure 6 expresses so many units 



NUMERATION 7 

of the third order, which is hundreds, and as in this case 
the figure is 3, three hundred is represented, and the 
above reads three hundred forty-six. 

The fourth figure of a number expresses so many 
thousands. 

The fifth expresses so many tens of thousands. 

The sixth expresses so many hundreds of thousands, etc. 

The nought (0), usually called cypher, signifies nothing 
when standing alone, but when it stands with other 
figures it means something. 

For example, the cypher standing alone, thus, 0, 
means that no units are represented. 

And the figure 1 standing alone, thus, 1, denotes one 
unit only, but by connecting a cypher to the 1, thus, 10, 
the value of 1 is increased tenfold, making it ten because 
it is removed one place to the left. Again, 13 standing 
alone, thus, 13, denotes thirteen units, and by connecting 
a cypher to it, thus, 130, the figures 13 are removed 
one place to the left each and thereby given a tenfold 
value and become one hundred and thirty. 

Again, 103 (one hundred and three) ; here the cypher 
has a similar significance on the figure 1, causing it to 
represent one hundred. By referring to page 5 we will 
see that each three places of figures are separated by a 
comma (,) and each three places thus separated denote 
a group, and each group has a name as shown below: 

000, 000, 000, 000, 

Billions. Millions. Thousands. Hundreds. 

The first set of three on the right are hundreds, the 
next set of three are thousands, the next set of three are 
millions, and the next set of three are billions. 

With the aid of the following table, express in writing 
the value of the numbers shown: 



8 A HAND BOOK FOR MECHANICS 

02 

w 52 S 

.2 - 2 § -3 

1 1 M 1 1 

o =3 o =3 c 2 • 

173 •'- , J3 m ■ 

o> <** 03 <u«4-oq a)t4_2 a» 

iofl 45 o g ,£ c $ tj 

12 » .2 ^ <* -2 ^^2 ^ » -2 

S g — Sc=3 § = 2 § c *3 

ffiHPQ K H ^ ffi H H KHP 
0, 0, 0, 0, 

--- --5 Number 1 

-25 Number 2 

342 Number 3 

1 674 Number 4 

-54 340 Number 5 

960 780 Number 6 

--3 726 90 2 Number 7 

--- -15 908 660 Number 8 

302 607 606 Number 9 

--5 003 000 003 Number 10 

-32 673 300 300 Number 11 

999 999 999 999 Number 12 

Example. Read, 463463765322. 

First begin and mark off the figures in threes from 
right to left; thus, 

463,463,765,322 

and the number reads four hundred and sixty three 
billion four hundred sixty-three million seven hundred 
and sixty-five thousand three hundred and twenty-two. 

Example. Read, 1070008346. 

First mark off in threes from right to left as in last 
example; thus, 

1,070,008,346 



NUMERATION 9 

and the number reads one billion seventy million eight 
thousand three hundred and forty-six. 

Examples for Exercise 
Read the following numbers: 



1. 


4364 


2. 


1927 


3. 


9009 


4. 


434672 


5. 


6497923 


6. 


53290678 


7. 


500490069 


8. 


5867340068 




NOTATION 



Notation is the art of writing or otherwise expressing 
numbers by means of figures. 

Example. Express in figures three hundred and 
twenty-two. Ans. 322. 

Example. Express in figures three thousand two 
hundred and twenty-three. Ans. 3223. 

It is very difficult for those who have had little practice 
in writing down numbers, to do so correctly; for such 
the following is recommended, by means of which the 
art of notation will be made very simple. 

Place in a row a number of cyphers (0) and mark 
them off in threes from right to left with a comma(,); 
and over the tops of each set of three (groups) note the 
positions of hundreds, thousands, millions, etc. Thus: 

Billions. Millions. Thousands. Hundreds. 

















8 










18 





5 



10 A HAND BOOK FOR MECHANICS 

Now with the aid of the above express in figures eight 
million. This is done by using the table as a guide and 
placing eight million under its proper position as shown 

Again express in figures, with the aid of the above, 
eighteen million and fifty. 

This is done by placing eighteen million under its 
proper position place next the fifty; then fill out the 
spaces between these figures with cyphers, as above, 
and it stands 18,000,050, which is the correct expression. 

Examples for Exercise 

1. Express in figures three thousand seven hundred 
and twenty-five. 

2. Express in figures twelve thousand six hundred. 

3. Express in figures three hundred and twenty-two 
thousand and six. 

4. Express in figures six million fifty thousand and 
twenty. 

5. Express in figures thirty-eight million four hundred 
and twenty thousand three hundred and fifty. 

6. Express in figures two hundred and twenty-two 
million eight hundred and thirty-five thousand one 
hundred and thirty. 

ADDITION 

Addition is the art of expressing in one number the 
units contained in two or more numbers. Making a 
single number, for instance, express the number of units 
contained in the numbers 342 and 721. 

In addition care must be taken in placing the figures. 
The unit figure of one line must be placed under the unit 
figure of the line above it. The same care must be taken 
in placing the tens, hundreds, thousands, etc. That is, 
the unit figure in the second line of figures must be 



ADDITION 11 

placed directly under the unit figure of the first line of 
figures, and the unit figure of the third line under the 
unit figure of the second line, etc. 

The same rule must be followed in placing the tens, hun- 
dreds, thousands, etc. The tens being placed under the 
tens of the line of figures above it, likewise the hundreds 
must be placed under the hundreds of the line above; 
also the thousands must be placed directly under the 
thousands of the line above. It does not matter if there 
are more figures in one line of figures than in another line. 

For instance, one line having three figures can be added 
to a line having four or any number of figures. And 
the lines of figures in such a sum would be placed thus: 

s "& » « 

H W H P 

7 6 3 2 Line having four figures. 
14 2 Line having three figures. 

Placing the unit figure of the second line under the 
unit figure of the first. The tens and hundreds of the 
second line are likewise placed under the tens and hun- 
dreds of the first line as shown. 

The following examples will show more fully how the 
figures in each line are placed: 



6342 


1 


87654216 


42 


32 


439103 


791 


344 


716724332 


1002 


6778 


4986345 



And it will be seen thereby, as just explained, that 
units are placed under units, tens under tens, hundreds 
under hundreds, etc. 



12 A HAND BOOK FOR MECHANICS 

Rules for Adding 

Add the right-hand column first and write the unit 
figure of the sum of the column under it, and add the 
tens, if any, to the next column. So proceed with all 
the columns, writing down the unit figure only of the 
sum of each column, with the exception of the last 
column, in which case the entire sum of the column is 
written down. 

Example. Add together 43 and 52 and 16 and 8. 
Proceed by placing each row of figures one under the 
other, as follows: 

Thus, 43 

52 

16 

8 

119 

Having arranged the numbers as in the example, we 
first add the units, which is the right-hand column, and 
we say 8 and 6 are 14 and 2 are 16 and 3 are 19; now 
in 19 units there are 9 units and 1 ten, so we write the 
9 units under the column of units and add the 1 ten to 
the column of tens, then 1 added to 1 makes 2, and 2 
and 5 are 7 and 4 are 11, and in this case the sum of 
the whole column is written down because it is the last 
column to be added, and we find when the four numbers 
are added together they equal the sum of 119. 

Example. Add together the following: 346, 794, 19, 

and 6 ' 346 

794 
19 

6 

1165 



ADDITION 13 

Here proceed as before, first by placing the different 
numbers to be added together in their proper relative 
places. 

Then add first the units, thus, 6 and 9 are 15 and 4 
are 19 and 6 are 25. Write the 5 units under the 
column of units and add the 2 tens to the column of tens, 
thus, 2 and 1 are 3 and 9 are 12 and 4 are 16. Write 
the 6 tens under the column of tens and add the 1 
hundred to the next column, then 1 and 7 are 8 and 3 
are 11; then the sum of this column equals 11 hundreds 
or one thousand and 1 hundred, in which case we place 
the 1 hundred under the column of hundreds, and as 
there is no other column to be added, which, if there 
was, would be thousands, we place the one thousand in 
thousands place, and we find the sum of the four numbers, 
when added together, equals 1165. 

Examples. Add together: 



643 


346 


421 


752 


132 


3 


16 


684 


24 


8 


2 


342 


1419 .4ns. 


1164 Ans. 


790 



When adding columns of figures where the sums of 
the columns will equal three figures or more, the right- 
hand figure of the sum only is written down, under the 
column, and the other two figures are added to the next 
column, as in the above example, where the columns 
summed up to two figures, the unit figure only was writ- 
ten down and the tens added to the next column. 

For example, if it is found that the sum of the column 
of figures equals 342, the right-hand figure 2 only is put 
down, and the 34 is added to the next column, whose 



14 



A HAND BOOK FOR MECHANICS 



sum we will suppose equals 1436, with the addition of 
the 34. Then the right-hand figure 6 of the sum of the 
column is written down only, and the 143 is added to 
the next column, which, if it happened to be the last 
column, would have its whole sum written under it. 

Examples for Practice 



Add together: 










(1) 


(2) 




(3) 


(4) 


3421 


7996 


6754 


694004 


6724 


8432 


4576 


426772 


5431 


6013 


5467 


324584 


2031 


0421 


9100 


953267 


17607 


22862 


25897 


2398627 


(5) 




(6) 




(7) 


3421 




7899 




8998 


7996 




9889 




5496 


6724 




8998 




8448 


6999 




5496 




9779 


8768 




8448 




8889 


9999 




7997 




7899 


8888 




9888 




9889 


7777 




6989 




9896 


5989 




8888 




8888 


9898 




9999 




9889 


7987 




9999 




8998 


6893 




8768 




9999 


5496 




6999 




8768 


8998 




8626 




7688 


9889 




7997 




8876 


7899 




3421 




8768 



123621 



130301 



141168 



ADDITION 15 

Examples for Exercise 

Add together: 

1. Thirty-two thousand seven hundred and two; six- 
teen thousand four hundred and sixty-seven; nine thou- 
sand three hundred and forty-four; forty-two thousand 
one hundred and sixty-four. 

2. One million three hundred and sixty-seven thousand 
four hundred and twenty-two; five million four hundred 
thousand; three hundred and sixty-three; one thousand 
three hundred and forty-two; seven million one hundred 
and sixteen thousand and forty-two ; four million eighteen 
thousand seven hundred and fifty-three. 

3. Six hundred and ninety-four thousand and four; 
four hundred and twenty-six thousand seven hundred 
and seventy-two ; three hundred and twenty-four thousand 
five hundred and eighty-four; nine hundred and fifty- 
three thousand two hundred and sixty-seven. 

4. Five million four hundred and sixty thousand seven 
hundred and twenty; seven million thirty-two thousand 
and thirty; two million three hundred and seventy-two 
thousand four hundred and fifty; one hundred and twelve 
million three hundred and sixty-two thousand nine 
hundred and seventy-two. 

5. Seven hundred and twenty-two million three hun- 
dred and thirty-three thousand five hundred and sixty- 
seven; two million thirty thousand and thirty; seventeen 
million seven hundred and twenty thousand six hundred 
and seventy-two; forty-two million five hundred and 
eighty-two thousand three hundred and thirty-six. 

Find the sum of: 

1. 4627 and 9824 and 2763 and 9005. 

2. 916742 and 7104 and 83 and 7628143. 

3. 18926742 and 6742839 and 15078949 and 1672. 



16 A HAND BOOK FOR MECHANICS 

SUBTRACTION 

Subtraction is the art or process of taking one number 
or quantity from another number or quantity of the same 
order, or the operation of finding the difference between 
two numbers of the same kind. For example, the process 
of finding the difference between the numbers 1674 and 
4236. 

Like denominations may be subtracted one from the 
other only. That is, 25 cents can be taken from 75 cents, 
but 25 cents cannot be taken from 75 potatoes. Neither 
can 36 bushels of wheat be taken from 96 bushels of oats, 
but 36 bushels of oats may be taken from 96 bushels of 
oats. 

That number or quantity to be subtracted from is 
called the minuend, and that number or quantity which 
is taken is called the subtrahend. And the result of the 
operation of taking one number from another is called 
the difference. 

Minuend means to be diminished. 

Subtrahend means that number which is taken from 
another. 

Only units of the same order can be subtracted. That 
is, ones can be taken from ones; tens from tens; hundreds 
from hundreds, etc. 

Example. Find the difference between 684 and 342. 

Proceed thus: 

First place the subtrahend 342 under the minuend 
684, arranging the figures in columns so that units come 
under units, tens under tens, and hundreds under hun- 
dreds. As follows: 

684 Minuend 
342 Subtrahend 
342 



SUBTRACTION 17 



Then draw a line underneath, thus 

Proceed now to subtract, beginning with the column 
on the right, saying 2 ones from 4 ones leaves 2 ones, 
which write beneath in ones " place. So proceed with 
tens, saying 4 tens from 8 tens leaves 4 tens, which write 
under tens place; likewise proceed with the hundreds 
and say, 3 hundreds from 6 hundreds leaves 3 hundreds, 
which write under the hundreds place, and the difference 
is 342 as shown in example. 

Example. Subtract 264 from 498. 

Here proceed as before, first by placing 264, the sub- 
trahend, under 498, the minuend. Thus: 

498 Minuend 
264 Subtrahend 



234 Difference 

Then say, 4 ones from 8 ones leaves 4 ones, which 
write under the ones place; then 6 tens from 9 tens 
leaves 3 tens, which write under the tens place; then 
2 hundreds from 4 hundreds leaves 2 hundreds, which 
write under the hundreds place, and the difference is 234. 

The work of subtraction may be proven by adding 
together the subtrahend and the difference, and if the 
sum thus obtained is equal to the minuend the operation 
has been correctly performed. 

Thus, to prove the work of the above add together 
264, the subtrahend, and 234, the difference, and the sum 
498 equals the minuend. 

Examples for Exercise 

1. Subtract 462 from 896. Ans. 434. 

2. Subtract 256 from 467. Ans. 211. 

3. Subtract 372 from 985. Ans. 613. 



18 A HAND BOOK FOR MECHANICS 

4. Subtract 137 from 249. Ans. 112. 

5. Subtract 283 from 999. Ans. 716. 

6. Subtract 567 from 878. Ans. 311. 

7. If six apples cost 18 cents and 7 oranges cost 6 
cents, how many more cents did the apples cost than the 
oranges. Ans. 12. 

8. A boy had 46 marbles and he gave away 6 to one 
friend, 9 to another; how many marbles had he left? 
Ans. 31. 

9. John is 37 years old and Henry is 16, what is the 
difference in the ages? Ans. 21. 

10. A man had 75 dollars and he spent 24; how many 
dollars had he left? Ans. 51. 

11. Bought 100 pounds of iron for 90 dollars and sold 
it for 60 dollars; how much was lost on the transaction? 
Ans. 30. 

If any order of the minuend has less units than the 
same order of the subtrahend, increase its units by ten, 
and subtract ; consider the unit of the next minuend order 
one less and proceed as before. In other words, when it 
is found that the figure in the minuend to be taken from 
is less than the figure in the subtrahend, which is to be 
taken from it, increase then the units of that figure of 
the minuend ten times, and the next figure in the minuend 
to be subtracted from is decreased one unit. 

For instance: From 37 
Take 18 

19 

From 7 ones we cannot take 8 ones, so 7 is increased 
by 10 units and becomes 17 units. We now say 8 units 
from 17 units leaves 9 units, which write down. Now 
the next figure 3 is decreased by one unit, so becomes 2; 
then we say 1 ten from 2 tens leaves 1 ten, which write 



SUBTRACTION 19 

down, and the difference between 37 and 18 is 19, as shown 
in example. 

Example. From 566 
Take 388 



178 

Here as in the previous example, the unit 8 in the 
subtrahend is larger than the unit 6 in the minuend, so 
we add 10 to the 6 and consider it 16 units; then say 
8 units from 16 units leaves 8 units, which write under- 
neath. Now the next figure 6 is to be considered 5, 
because, as just explained, 1 unit must be taken from 
this unit when the unit before it is increased by 10 units; 
it then becomes 5 instead of 6. Now 8 tens from 5 tens 
cannot be taken, so w T e increase the 5 tens by 10, and 
it then becomes 15 tens; so we say 8 tens from 15 tens 
leaves 7, which write underneath. Now the next order 
5 is to be considered 1 less for the same reason that 
the last order 6 was considered 5; then say, 3 hundreds 
from 4 hundreds leaves 1 hundred, which write down, 
and the difference is found to be 178 as shown in the 
example. 

Example. From 6605 
Take 3767 



2838 



Seven units cannot be taken from 5 units, so 10 units 
are added to the 5 units, making it 15 units; then 
7 units from 15 units leaves 8 units, which write under- 
neath. Now when the next order to be subtracted from 
is 0, as in the example, 10 units are likewise added, 
whereupon the would be increased to 10 tens. Now 
from these 10 tens 1 unit is taken, because the unit 
before it was increased 10 units, therefore it becomes 9; 



20 A HAND BOOK FOR MECHANICS 

then we say, 6 tens from 9 tens leaves 3 tens, which 
write underneath the tens place. Then proceed by 
taking one unit from the next order, making it 5, and as 
7 cannot be taken from 5 we add 10 units thereto, making 
it 15; then say 7 from 15 leaves 8, which write under- 
neath. Then from the next order 6 take 1 unit, which 
makes it 5; then say 3 from 5 leaves 2, which write 
underneath, and the difference is 2838, as shown in 
example. 

Examples for Exercise 

1. From 397 take 279. Ans. 118. 

2. From 966 take 788. Ans. 178. 

3. From 4592 take 2364. Ans. 2228. 

4. From 6422 take 3188. Ans. 3234. 

5. From 9263 take 6346. Ans. 2917. 

6. From 8674 take 5785. Ans. 2889. 

When any order of the minuend has the same number 
of units as the same order of the subtrahend, its units 
are not increased by tens, as was the case in the pre- 
ceding examples in which the order of the units in the 
minuend were less than the units of the same order in 
the subtrahend. 

In other words, when the figure to be taken from is 
the same as the figure which is to be taken from it, no 
matter what the order is, of the figures which are to be 
subtracted one from the other, the figure to be subtracted 
from is unchanged. 



■&' 



For example: From 14 
Take 4 

10 

In this case the unit figures in the minuend and sub- 
trahend, are the same, in which case we say 4 ones from 



SUBTRACTION 21 

4 ones leaves ones, which write down in units or ones 
place. Then there are no tens to be taken from the 
1 ten, so we say ten from 1 ten leaves 1 ten, which 
write underneath in tens place, and the answer is 10 as 
shown in example. 

Example. From 654 

Take 354 

300 

Here is an example having two figures in the subtra- 
hend, the same as two figures of the same order in the 
minuend. 

So proceed as in last example by taking 4 units from 
4 units, which leaves units, which write underneath in 
units or ones place; then say 5 tens from 5 tens leaves 
tens, which write underneath tens place; then say 
3 hundreds from 6 hundreds leaves 3 hundreds, which 
write underneath hundreds place, and the answer is 300., 
as shown in example. 

Example. From 767 

Take 558 



209 



In this case the unit figure 7 to be subtracted from is 
less than the unit figure 8 which is to be subtracted 
from it, so increase it by 10 units, and it becomes 17 
units, from which take the 8 units in the subtrahend and 
9 units are left, which write underneath in ones place. 
Now the next order of the minuend, which is 6, is to be 
considered one less, which makes it 5; then say 5 tens 
from 5 tens leaves tens, which write underneath in 
tens place; then say 5 hundreds from 7 hundreds leaves 
2 hundreds, which write down underneath hundreds place, 
and the answer is 209 as shown in example. 



22 A HAND BOOK FOR MECHANICS 

Example: What is the difference between 
89645278 
and 65437169 



24208109 Arts. 

Examples for Exercise 

1. What is the difference between 486795 and 245483. 
Ans. 241312. 

2. What is the difference between 937256 and 518137. 
Ans. 419119. 

3. What is the difference between 76042912 and 537- 
96878. Ans. 2246034. 

4. What is the difference between 843676 and 796832. 
Ans. 46844. 

5. What is the difference between 26784263 and 
25342621. Ans. 1441642. 

6. What is the difference between 7500476 and 456- 
7543. Ans. 2932933. 

MULTIPLICATION 

Multiplication Table 



1 i 2 


3 ! 4 | 5 


6 
12 

18 


7 | 8 


9 


10 


11 


12 


2 4 


6 j 8 1 10 


14 | 16 


18 


20 


22 


24 


3 6 


9 | 12 15 


21 | 24 


27 


30 


33 


36 


4 | 8 


12 | 16 | 20 


24 


28 j 32 


36 


40 


44 


48 


5 ; 10 : 15 J 20 J 25 


30 


35 i 40 


45 


50 


55 


60 


6 ! 12 ! 18 i 24 1 30 

1 i 


36 


42 j 48 


54 


60 


66 


72 


7 , 14 | 21 | 28 ! 35 


42 


49 ! 56 


63 


70 


77 


84 


8 j 16 | 24 ! 32 | 40 


48 


56 | 64 


72 


80 


88 


96 


9 | 18 | 27 | 36 i 45 


54 | 63 i 72 


81 


90 


99 


108 


10 20 , 30 40 50 


60 i 70 ; 80 


90 


100 


110 


,120 


11 22 j 33 44 j 55 


66. ! 77 : 88 


99 


110 


121 


134 


12 24 36 i 48 60 72 84 96 


108 


120 


132 


144 



MULTIPLICATION 23 

Multiplication is a convenient method of finding the 
sum of any number that has to be added to itself any 
number of times. 

For instance, if we had occasion to find the sum of 
6 added to itself 5 times. 

Instead of performing the operation thus: 

6 

6 

6 

6 
J5 
30 

we perform the operation by the process of multiplica- 
tion. The 6 is written down once only, and underneath 
it the 5, underneath which draw a line as shown in the 

following : „ 

6 

_5 

30 

Proceed then, saying 5 times 6 are 30, which place 
underneath the line as shown. 

It is here seen that both methods give the same results, 
but that the latter is the more convenient. 

Multiplication, then, is the process of finding the 
amount of one number increased as many times as there 
are units in another. 

In multiplication there are three terms used called the 
Multiplicand, Multiplier, and the Product. 

The Multiplicand is the number to be multiplied. 

The Multiplier is the number by which we multiply. 

The Product is the answer. 

According then to multiplication, one number (called 
the multiplier) is made to operate on another number 



24 A HAND BOOK FOR MECHANICS 

(called the multiplicand) as many times as there are 
units in the multiplier, and the result of the operation is 
called the product. As shown in the following example, 
where 7, the multiplier (the number by which we mul- 
tiply), operates on 8, the multiplicand (the number to be 
multiplied), and the result of the operation, 56, is the 
product. 

8 Multiplicand 

7 Multiplier 

56 Product 

Always multiply from right to left. 

When arranging the numbers to be multiplied one by 
the other, care must be taken to place the unit figure of 
the multiplier under the unit figure of the multiplicand. 
The same rule must be carried out in placing the tens, 
hundreds, thousands, etc.; that is, tens under tens, 
hundreds under hundreds, and thousands under thou- 
sands. 

Example. Multiply 34231 by 2. 

Proceed as follows: Write down the multiplicand 34231, 
under the unit figure 1 of which place the multiplier 2, 
underneath which draw a line as shown. 

34231 Multiplicand 
2 Multiplier 

68462 Product 

Then after arranging the sum thus, proceed to multi- 
ply from right to left, the figures in the multiplicand by 
the figure in the multiplier, saying 2 times 1 are 2, which 
set down underneath the 1 ; then say 2 times 3 are 6, 
which set down underneath the 3; then say 2 times 2 
are 4, which set down underneath the 2; then say 2 



M ULTIPLICA TION 25 

times 4 are 8,, which set down underneath the 4; then 
say 2 times 3 are 6, which set down underneath the 3. 

Examples for Practice 

Multiply 34213 4324312 

By 3 2 

102639 8648624 

When numbers are multiplied one by the other, the 
unit figure only of each product is set down, and the 
tens are added to the product of the next figure to the 
left, with the exception of the last figure to be multiplied, 
in which case the product in full is set down. 

Example. Multiply 642 by 3. 

Here, as in the preceding example, arrange the sum, 
placing the multiplier 3 under the unit figure 2 of the 
multiplicand, and drawing a line underneath. Thus: 

642 
3 

1926 

Proceed by multiplying the 2 of the multiplicand by 
the multiplier 3, saying 3 times 2 are 6, which set down 
directly under the 3; then multiply the 4 in the multi- 
plicand by 3, saying 3 times 4 are 12, but set down the 
unit or right-hand figure 2 only underneath the 4, and 
add the left-hand figure of the product, which is 1, to 
the next product. We then proceed, multiplying the 
6 in the multiplicand by the multiplier 3, saying 3 times 
6 are 18, to which, as before stated, 1 is to be added, 
making it 19, which is set down in full, because it is the 
product of the last figure to be multiplied. 

Example. Multiply 64395 by 4. 



26 A HAND BOOK FOR MECHANICS 

Proceed as before by placing the multiplier under the 
unit figure of the multiplicand, drawing a line under- 
neath. 64395 

4 



257580 



Then multiply the 5 in the multiplicand by 4, saying 
4 times 5 are 20, and set down the (cypher) directly 
underneath the 4. We then multiply the 9 in the multi- 
plicand, saying 4 times 9 are 36, to which add the 2 tens 
of the last product, making 38; setting down the 8 only 
of this product, we add the 3 to the next product; saying 
4 times 3 are 12 and 3 are 15, we set down the 5 only 
and add the 1 to the next product and proceed, saying 
4 times 4 are 16 and 1 are 17, the right hand or unit 
figure of which set down only, and add the 1 to the next 
product, saying 4 times 6 are 24 and 1 are 25, which, as 
it completes the multiplication we set down in full and 
the product is 257580, as shown. 

Examples for Practice 



Multiply 132 
By 3 


432 
2 


212 
4 


396 


864 


848 


Multiply 634 
By 2 


732 
3 


845 
5 


1268 


2196 


4225 


Multiply 
By 


9654321672 

7 






67580251704 





MULTIPLICATION 27 

Example. Multiply 4635 by 11. 

Here is an example having two figures in the multiplier, 
in which case, when the multiplier does not exceed 12, 
the operation is the same as in the preceding examples. 

4635 
11 



50985 



Having arranged the numbers to be multiplied one by 
the other, placing the unit figure of the multiplier under 
the unit figure of the multiplicand, and the tens of the 
multiplier under the tens of the multiplicand, we multiply 
the 5 in the multiplicand by 11, saying 11 times 5 are 55; 
set down the right-hand figure only. We then say, 
11 times 3 are 33, to which the left-hand figure 5 of the 
preceding product is added, making 38; we set down the 

8 under the 3 of the multiplicand. Now multiply 6 by 
11, saying 11 times 6 are 66 and 3 are 69, the 3 being the 
left-hand figure of the preceding product; set down the 

9 only of the product and add the 6 to the next product. 
Now multiply 4 by 11, saying 11 times 4 are 44 and 6 
are 50, which set down in full as it finishes the multipli- 
cation, and the product is 50985. 

Example. Multiply 6324 by 12 

12 



Set down the multiplicand and the multiplier as before, 
placing the unit figure of the multiplier under the unit 
figure of the multiplicand, and the tens of the multiplier 
under the tens place of the multiplicand. Thus: 

6324 
12 

75888 



23 A HAND BOOK FOR MECHANICS 

Then proceed to multiply each figure of the multipli- 
cand by the multiplier, saying 12 times 4 are 48, of 
which we set down the right-hand figure 8 only. We then 
multiply the 2 in the multiplicand by 12, saying 12 times 
2 are 24, to which 4, the left-hand figure of the pre- 
ceding product, is added, making 28; set down the 8 
only. We then multiply 3 by 12, saying 12 times 3 are 
36 and 2 (the left-hand figure of preceding product) 
are 38; set down the 8 only of this product. We then 
multiply the 6 in the multiplicand by 12, saying 12 
times 6 are 72 and the 3 of the preceding product are 75, 
which is the product of the last figure to be multiplied 
and is set down in full. 

The whole product then is 75888. 

Examples for Practice 

Multiply 52436 
By 10 

524360 

Multiply 63425 
By 12 



761100 
Multiply 724302 
By 11 

7967322 

When the multiplier consists of more than one figure 
(and exceeds 12), each figure of the multiplicand is 
multiplied by each figure of the multiplier. 

For instance, if we had occasion to multiply 463 by 
24, in which case the multiplier exceeds 12, we would 
proceed after arranging the numbers to be multiplied, 
thus: 



MULTIPLICATION 29 

463 
24 



1852 
926 
11112 



by multiplying each figure of the multiplicand 463 by 
the unit figure 4 of the multiplier first. Then in like 
manner each figure of the multiplicand would be multi- 
plied by the 2 of the multiplier. 

The products obtained by the operation of each figure 
of the multiplier upon the multiplicand are called partial 
products, and the right-hand figure of each partial 
product is placed directly under the figure of the multiplier 
that produces it. 

Thus, in the above example, we first multiplied each 
of the figures 463 of the multiplicand by the figure 4 
of the multiplier, and by the process got the partial 
product 1852, the unit figure 2 of which is placed directly 
under the figure 4 of the multiplier. 

We then multiplied each of the figures 463 by 2 and 
get a second partial product 926, the unit figure 6 of 
which is placed directly under the figure 2 of the multi- 
plicand which produced it. 

To get the whole product add together the several 
partial products. 

Example. Multiply 433 by 32. Write the multiplier 
and multiplicand as before. 

433 
32 



866 
1299 

13856 



30 A HAND BOOK FOR MECHANICS 

Proceed to multiply each figure of the multiplicand 
by each figure of the multiplier successively. Beginning 
with the unit figure 2 of the multiplier we multiply each 
figure of the multiplicand, saying 2 times 3 are 6, which 
set down underneath the multiplier 2. Then multiply 
the next figure 3 of the multiplicand by the unit figure 

2 of the multiplier, saying 2 times 3 are 6, which set 
down. Then multiply the figure 4 of the multiplicand 
by the unit figure 2 of the multiplier, saying 2 times 4 
are 8, which set down. 

By the operation of the figure 2 of the multiplier upon 
each figure of the multiplicand, the partial product 866 
is obtained. 

Proceed now to multiply each figure of the multiplicand 
by the next figure 3 of the multiplier, saying 3 times 3 
are 9, and, as before stated, the right-hand figure of each 
partial product is placed directly under the figure of the 
multiplier which produces it; the 9 is placed under the 

3 of the multiplier. 

Now multiply the next figure 3 of the multiplicand by 
the figure 3 of the multiplier, saying 3 times 3 are 9, 
which set down underneath the next figure to the left 
of the preceding partial product. Now we multiply the 
next figure 4 of the multiplicand by the figure 3 of the 
multiplier, saying 3 times 4 are 12, which set down in 
full, and by the operation of the figure 3 of the multiplier 
upon each figure of the multiplicand we have obtained 
another partial product, 1299, underneath which, as it 
is the last partial product to be obtained, we draw a 
line and proceed to get the whole product by adding 
the partial products together as in addition, and we have 
for the whole product, 13856, as shown. 

Example. Multiply 543 by 43. 



MULTIPLICATION 31 

Here arrange the numbers to be multiplied one by 
the other as before. 

543 
43 



1629 
2172 

23349 



And this being another example where the multiplier 
exceeds 12, we proceed as before to get the product by 
making each figure of the multiplier operate upon each 
figure of the multiplicand successively. Beginning with 
the figure 3 of the multiplier, first we multiply the 3 of 
the multiplicand, saying 3 times 3 are 9, which set down 
directly underneath the 3 of the multiplier, which pro- 
duced it; then multiply the next figure of the multipli- 
cand by the figure 3 of the multiplier, saying 3 times 4 
are 12; of this product we set down the 2 and add the 
1 to the next product and proceed, saying 5 times 3 are 
15 and 1 are 16, which set down in full. 

Now proceed to multiply by the next figure 4 of the 
multiplier each figure of the multiplicand, saying 4 times 

3 are 12; here set down the right-hand figure 2 under 
the figure 4 which produced it, and add to the next 
product the left-hand figure 1 and proceed, saying 4 
times 4 are 16 and 1 are 17; set down the 7 only under 
the next place to the left of the preceding partial product 
1629, and to the next product add the 1; proceed, saying 

4 times 5 are 20 and 1 are 21, which set down in full. 
Now having obtained the partial products, draw a line 

underneath as before and proceed to add them together 
to get the whole product, and as shown the product of 
543 multiplied by 43 is 23349. 



32 A HAXD BOOK FOR MECHANICS 

Examples for Practice 



Multiply 
By 


524 
34 


467 
53 




2096 
1572 


1401 
2335 


Multiply 
By 


17816 

693 

91 


24751 

396 

82 




693 
6237 


792 

3168 



63063 32472 

Example. Multiply 5943 by 434. 

Here is an example having three figures in the multi- 

Plier ' 5943 

434 



23772 

17829 
23772 

2579262 



After arranging the multiplier under the multiplicand, 
placing units under units, tens under tens, and hundreds 
under hundreds, proceed to multiply each figure of the 
multiplicand by each figure of the multiplier, as in pre- 
ceding examples, placing the right-hand figure of each 
of the three partial products directly under the figure of 
the multiplier which produces it. That is, place the 
right-hand figure obtained by the operation of the unit 
figure 4 of the multiplier upon each figure of the 
multiplicand directly under the 4. Place the right-hand 



MULTIPLICATION 33 

figure of the partial product obtained by the operation 
of the figure 3 of the multiplier upon each figure of the 
multiplicand under the figure 3 of the multiplier which 
produces it, and place the right-hand figure of the partial 
product obtained by the operation of the figure 4 of the 
multiplier upon each figure of the multiplicand directly 
under the figure 4 of the multiplier which produces it. 

Having obtained the partial products, add them to- 
gether and we have the whole product of 5943 multiplied 
by 434 equal to 2579262. 

When numbers are multiplied one by the other, the 
product is the same whatever the order of multiplying 
them. That is, the product of 642 multiplied by 64 is 
the same as the product of 64 multiplied by 642, as 
shown by the following examples: 

642 64 

64 642 

2568 128 

3852 256 

41088 384 



41088 



For convenience, however, it is customary to multiply 
the larger number by the smaller. 

Method of Operation where there are Cyphers 
in the Multiplier. 

Example. Multiply 5634 by 403. 

Here we multiply by the figures of the multiplier only, 
placing the right-hand or unit figure of each partial 
product directly under the figure of the multiplier which 
produces it, and pass over the cypher or cyphers in the 
following manner: 



34 A HAND BOOK FOR MECHANICS 



After arranging 
the other, thus: 


the numbers to be 
5634 


multiplied 


one 


by 






403 










16902 
22536 





2270502 

we proceed to multiply first by the figure 3 of the 
multiplier and obtain by the operation the partial product 
16902. 

Now we come to the cypher (0), which we pass over 
and multiply by the next figure 4, taking care to place 
the right-hand figure 6 of the partial product 22536, 
obtained by the multiplication, directly under the figure 
4 of the multiplier which produces it. 

Now having obtained the partial products, add same 
together and the whole product is seen to be 2270502. 

Example. Multiply 9436782 by 24003. 

Arrange the numbers to be multiplied one by the 
other as before. Thus: 

9436782 
24003 



28310346 
37747128 
18873564 

226511078346 



Proceed now to multiply the figures of the multiplicand 
by the figure 3 of the multiplier, by which operation is 
obtained the partial product 28310346, the right-hand 
figure 9 of which is placed directly under the 3 of the 
multiplier which produces it. We now come to the 



MULTIPLICATION 35 

cyphers of the multiplier, two of them, which we pass 
over and use the figure 4 as the next multiplier. By the 
operation of the figure 4 upon the multiplicand, the 
partial product 37747128 is obtained, the right-hand 
figure 8 of which place directly under the figure 4 of the 
multiplier which produces it. Proceed now to multiply 
each figure of the multiplicand by the figure 2 of the 
multiplier, by which operation is obtained the partial 
product 18873564, the right-hand figure 4 of which place 
directly under the 2 of the multiplier which produces 
it. 

Having now obtained the several partial products, add 
them together and we have the whole product as shown, 
226511078346. 

Examples for Practice 



Multiply 
By 


4532 
203 


2354 
303 




13596 
9064 


7062 
7062 


Multiply 
By 


919996 

564672 
23004 


713262 

679423657 
4000003 




2258688 
1694016 
129344 

2989714688 


2038270971 
2717694628 


1 
1 


2717696666270971 



Example. Multiply 425326 by 5300. 

Here is an example having cyphers (0) on the right of 
the multiplier. In which case the first significant figure 
of the multiplier (from right to left) is placed directly 



36 A if AM) hook FOR MECHANICS 

under the first figure of the multiplicand from right to 
left. 

Multiply by the significant figures only and annex as 
many cyphers to the product as there are cyphers on 
the right of the multiplier. Proceed now to multiply, 
arranging the first figure 3 of the multiplier under the 
first figure 6 of the multiplicand. Thus: 

425326 

5300 

1275978 
2126630 

2254227800 

Passover the cyphers and, as before stated, multiply 
by the significant figures only. 

First multiply the multiplicand 425326 by the figure 
3 of the multiplier 5300, by which operation we obtain 
the partial product L275978, the right-hand figure 8 of 
which is placed directly under the figure 3 of the mul- 
< iplier which produced it. 

Multiply now the multiplicand by the next figure 5 of 
the multiplier, by which operation is obtained the partial 
product, 2126630, the cypher C0j of which is placed 
directly under the figure 5 of the multiplier which pro- 
duced it. Add together the partial products thus obtained, 
to which annex the cyphers (two of them) which are at 
the right of the multiplier, and we obtain the whole 
product 2254227800 as shown. 

Example. Multiply 53267 by 4700. 

Arrange I he numbers to be mull iplied as before, placing 
the first figure 7 of the multiplier 4700 under the first 
figure 7 of the multiplicand, and place the uext figure 4 
of the multiplier one place to the left. Thus: 



MULTIPLICATION 

53207 
4700 



37 



372809 
2 1 3008 

250354900 



Proceed now t.o multiply by the figure 7 of the multi- 
plier first, then by the figure 4. 

Add together the partial products thus obtained, to 
which annex the cyphers on the right of the multiplier, 
and by the operation we obtain the product 250354900 
:iown. 

Ex iMPLBS J OB hi' U 'J J' 



Multiply 


135072 


270354 


By 


40 
5420880 


030 




820002 






1058124 






171103020 


Multiply 


4508012 


70235407 


By 


7200 


30000 




0137881 


2377004010000 




31082501 






32800382100 




Multiply 


32807 J 


1703254 


By 


1 00 


1000000 




32807100 


1703254000000 



Process- When there are cyphers on the right of 
both multiplicand and multip I down the significant 

figures of the multiplier under the significant figure 
the multiplicand, placing the first (from right to left) 



38 A HAND BOOK FOR MECHANICS 

significant figure of the multiplier directly under the 
first (from right to left) significant figure of the multipli- 
cand. Annex to the product obtained by the operation 
of the multiplier upon the multiplicand as many cyphers 
as there are on the right of the multiplicand and multi- 
plier. 

Example. Multiply G4200 by 340. 

Arrange the multiplier 340 under the multiplicand 
64200, placing the figure 4 of the multiplier (which is 
the first figure from right to left) under the figure 2 of 
the multiplicand (which is the first figure of the multi- 
plicand, from right to left. Thus: 

64200 
340 

2568 
1926 



21828000 



Proceed now to multiply. 

Multiplying the figures 642 of the multiplicand by the 
figure 4 of the multiplier, we obtain the partial product 
2568, the right-hand figure 8 of which is placed directly 
under the figure 4 of the multiplier which produces it. 

Now multiply the figures 642 of the multiplicand by 
the figure 3 of the multiplier, by which operation is 
obtained the partial product 1926, the figure 6 of which 
is placed directly under the figure 3 of the multiplier 
which produces it. 

Proceed now by adding together the partial products 
obtained. By so doing we get 21828, to which, as before 
stated, the cyphers on the right of the multiplicand and 
multiplier are annexed. We then have, by annexing the 
cyphers, the whole product 21828000 as shown. 



MULTIPLICATION 39 

Example. Multiply 80246000 by 200400. 

Here proceed as before, placing the first significant 
figure 4 of the multiplier under the first figure 6 of the 
multiplicand. Thus: 802 46000 

200400 



320984 
160492 

16081298400000 



Proceed now to multiply 80246 of the multiplicand by 
the figures 4 and 2 of the multiplier. 

By the operation of the figure 4 of the multiplier upon 
the figures 80246 of the multiplicand is obtained the par- 
tial product 320984, the right-hand figure 4 of which is 
placed directly under the figure 4 of the multiplier which 
produced it. 

Next in the multiplier are two cyphers, which pass 
over in the operation. We then multiply by the figure 2, 
by which operation is obtained the partial product 
160492, the right-hand figure 2 of which is placed directly 
under the 2 of the multiplier which produced it. 

Now add together the partial products thus obtained, to 
which annex as before the cyphers on the right of the mul- 
tiplicand and multiplier, and as shown the whole product 
of 80246000 multiplied by 200400 is 16081298400000. 

Examples for Practice 

Multiply 4020300 

By 2309 

361827 
120609 
80406 

9282872700 



40 A HAND BOOK FOR MECHANICS 

Multiply 5003600000 3060402 
By 2063100 3002010000 

50036 3060402 

150108 6120804 
300216 9181206 



100072 9187357408020000 

10322927160000000 

Examples for Exercise 

1. 45673 by 12 4. 4200600 by 3002000 

2. 567346 by 342 5. 954673000 by 700730 

3. 760263 by 3604 

DIVISION 

Division is the process of dividing, or separating into 
parts. The process of finding the number of times one 
number is contained in " another number." Finding, 
for instance, the number of times 4 units are contained 
in 20 units. 

There are three terms used in division, called the 
Divisor, the Dividend, the Quotient. 

The Divisor is the name applied to that number by 
which we divide. 

The Dividend is the name applied to that number 
which is divided or separated into parts. 

The Quotient signifies the number of times the Divisor 
is contained in the Dividend. 

In arranging numbers to be divided one by the other 
the divisor is placed on the left of the dividend, and 
the divisor and dividend are separated by a curved 
line, thus, ). When the divisor does not exceed 12 the 
operation of dividing is performed by a process called 
''Short Division." 



DIVISION 41 

Example. Divide 8 by 4. 

Proceed as follows: 

Set down the 4, after which draw a curved line, on the 
right hand of which line place the figure 8, under which 
draw a line, thus: 

The example will then be set down as follows: 

Divisor 4)8 Dividend 
2 Quotient 

Proceed now to divide the figure 4 (the divisor) into 
the figure 8 (the dividend) , saying 4 into 8 goes. 2 times. 

The figure 2 thus obtained is the "quotient" and is 
placed under the dividend figure 8, being separated 
therefrom by a line as shown. 

Example. Divide 6 by 3. 

Arrange the divisor and dividend as before. Thus: 

Divisor 3)6 Dividend 
2 Quotient 

Proceed then to divide the dividend 6 by the divisor 
3, saying 3 into 6 goes 2 times. 

The 2 thus obtained is the quotient and is placed 
directly under the dividend figure 6. 

Always divide from left to right. 

Example. Divide 46 by 2. 

Arrange the divisor 2 and the dividend 46 as before. 
Thus: 

2 )46 
23 

Proceed to divide the divisor 2 into the left-hand 
figure 4 of the dividend first, saying 2 into 4 goes 2 times, 



42 A HAND BOOK FOR MECHANICS 

which set down underneath the 4. Then divide the next 
figure 6 of the dividend by the divisor 2, saying 2 into 
6 goes 3 times, which set down underneath the 6, and, 
as shown, the quotient thus obtained is 23, which signifies 
that 2 is contained in 46, 23 times. 

Example. Divide 462 by 2. 

Arrange the divisor and dividend as before. 

2 )462 
231 

Then proceed to divide. 

Taking the left-hand figure 4 of the dividend first, 
say 2 into 4 goes 2 times, which place under the 4. Now 
divide the next figure 6 of the dividend by the divisor 2, 
saying 2 into 6 goes 3 times; place the 3 under the 6. 
Then proceed to divide the next figure 2 of the dividend 
by the divisor 2, saying 2 into 2 goes 1 times, place the 
1 obtained under the figure 2, and by the operation is 
obtained the quotient 231, as shown. 

Examples for Exercise 

Divide 6482 by 2. Thus: 

2 )6482 
3241 

Example. Divide 36366 by 3. Thus: 

3 )36366 
12122 

If there is a remainder after dividing the first or any 
of the following figures of the dividend, the remainder 
is prefixed to (placed to the left of) the next figure of the 
dividend. 



DIVISION 43 

For instance, if there was occasion to divide 74 by 2, 
Thus: 

2 )74 
37 

the divisor 2 is contained in the first figure 7 of the 
dividend 3 times and 1 over. Place the 3 in the quotient 
under the 7 and place the 1 which was left over on the 
left of the next figure 4 of the dividend. We then say 
2 is contained in 14 seven times; place the 7 thus ob- 
tained in the quotient under the 4; we have then in the 
quotient 37 as shown, which shows when there is a re- 
mainder it is prefixed to the next figure of the dividend, 
and the divisor is made to operate in the number thus 
formed. 

Example. Divide 762 by 6. 

Arrange the example as before. Thus: 

6 )762 
127 

Proceed then to divide, saying 6 is contained in 7 
1 time and one over. 

Write the quotient figure 1 directly under the 7 and 
imagine the remainder 1 to be placed to the left of the 
next figure 6 of the dividend, thus forming the number 
16. Inquire then the number of times 6 the divisor is 
contained in 16, which is found to be 2 times and 4 over. 
Place the quotient figure 2 under the 6 of the dividend, 
and to the next figure 2 of the dividend the remainder 
4 is prefixed. We then have 42. 

Now divide 6, the divisor, into 42, saying 6 into 42 goes 
7 times; place the figure 7 under the figure 2 of the divi- 
dend, and, as shown, the quotient is 127. 



44 A HAND BOOK FOR MECHANICS 

Example. Divide 948 by 4. 

4 )948 
237 

Having arranged the numbers to be divided one by 
the other, proceed to divide, saying 4 is contained in 9 
2 times and 1 over. Place the quotient figure 2 under 
the dividend figure 9 and prefix to the next figure 4 of 
the dividend, the remainder 1 making 14. 

We now divide the divisor 4 into the number 14 thus 
obtained, saying 4 is contained in 14 3 times and 2 over. 
Place the quotient figure 3 under the figure 4 of the 
dividend, and to the next figure 8 of the dividend prefix 
the remainder 2, making 28. 

Divide now 4, the divisor, into 28, saying 4 into 28 
goes 7 times, which place underneath the dividend figure 
8 and which completes the operation. 

Examples for Exercise 



3)4674 


4)6256 


5)7065 


1558 


1564 


1413 


0)68592 




7)792582 


11432 




113226 



Process. When the left-hand figure of the dividend 
is less than the divisor, annex it to the next figure or 
figures of the dividend, if more than one figure is re- 
quired to allow the divisor to operate in it. 

The quotient figure is placed directly under the last 
figure, which is annexed to the left-hand figure of the 
dividend. 

Example. Divide 651 by 7. 



DIVISION 45 

Arrange the divisor 7 and dividend 651 as before. 
Thus: 

7 )651 
93 

Proceed then to divide. 

It is seen that the divisor 7 is greater than the left- 
hand figure 6 of the dividend, in which case the next 
figure 5 of the dividend is annexed to the 6, making 65. 
Divide now by the 7, saying 7 is contained in 65 9 times 
and 2 over. Place the quotient figure 9 directly under 
the figure 5 of the dividend, and prefix the remainder 2 
to the next figure 1 of the dividend. By so doing we 
have 21, in which 7 is contained 3 times. Place the 3 
under the 1 of the dividend and the division is completed. 

Example. Divide 6344 by 8. 

Arrange the numbers to be divided one by the other. 
Thus: 

8 )6344 
793 

The left-hand figure 6 of the dividend is less than^the 
divisor 8, so to it we annex the next figure 3 of the 
dividend, making 63. 

Inquire now how many times the divisor 8 is contained 
in 63, saying 8 goes into 63 7 times and 7 over (remainder). 

Place the quotient figure 7 under the 3 of the dividend, 
and prefix the remainder 7 to the next figure 4 of the 
dividend, making 74. 

Inquire now the number of times the divisor 8 is 
contained in 74, saying 8 into 74 goes 9 times and 2 over. 
Place the quotient figure 9 obtained under the figure 4 
of the dividend, and prefix the remainder 2 to the next 
figure of the dividend, by doing which is obtained 24. 



46 A HAND BOOK FOR MECHANICS 

Divide this number 24 by the divisor 8, saying 8 into 
24 goes 3 times. Place the quotient figure 3 under the 
4 and the division is completed. 

Example. Divide 11424 by 12. 

12 )11424 
952 

It is seen that the divisor 12 is greater than the two 
figures on the left of the dividend, so the next figure 4 
is annexed thereto, making 114. Proceed now to divide 
by the divisor 12, saying 12 into 114 goes 9 times and 
6 over. Place the quotient figure 9 directly under the 
last figure (4) which was annexed to the two figures 11 
of the dividend as the first figure of the quotient. By 
prefixing the remainder 6 to the next figure of the divi- 
dend, we have 62, which contains the divisor 12 5 times 
and 2 over. 

Place the quotient figure 5 under the 2 of the dividend, 
and to the next figure 3 of the dividend prefix the re- 
mainder 2, making 24. Divide now the number 24 by 
the divisor 12, saying 12 into 24 goes 2 times. Place 
the quotient figure 2 under the 4 and the division is 
completed. 

Examples for Exercise 
6 )3254214 7)23541 



542369 


3363 


8)659696 


9)82368 


82462 


9152 


12)1105462128 





92121844 
Process. When the divisor is greater than the figure 



DIVISION 47 

of the dividend to he divided, place in the quofieni a 
cypher (0) directly under the figure, then to proceed 
with the division this figure is prefixed to the next figure 
of the dividend. The divisor is then made to operate 
in the new number thus formed. 

Exa.mplj;. Divide 3624 by 3. 

Arrange the divisor and dividend as before 

3 )3624 
1208 

Proceed now to divide, saying 3, the divisor, is con- 
tained in the figure 3 of the dividend 1 time. 

Place the figure 1 in the quotient under the figure 3 
of the dividend. 

Then divide the next figure 6 of the dividend by the 
divisor 3, saying 3 is contained in 6 2 times. 

Place the quotient figure 2 obtained under the figure 
6 of the dividend. 

It will be observed that the next figure 2 of the divi- 
dend is less than the divisor 3. In which case place 
directly under the figure a cypher (0). Consider the 
figure 2, now, to be prefixed (placed before) to the next 
figure 4 of the dividend, making 24. 

Proceed now to make the divisor 3 operate in the new 
number 24 thus obtained, saying 3 into 24 goes 8 times. 

Place the quotient figure 8 under the dividend figure 
4 and the division is completed. 

Example. Divide 243648 by 12. 

12 )243648 
20304 

Proceed to divide as follows: 

Inquire the number of times the divisor 12 is contained 



48 A HAND BOOK FOR MECHANICS 

in the two left-hand figures 24 of the dividend, saying 
12 into 24 goes 2 times. 

Place the quotient figure 2 underneath the figure 4 
of the dividend. The next figure 3 of the dividend is 
less than the divisor 12, in which case say 12 into 3 
goes times. Place the cypher (0) thus obtained di- 
rectly under the figure 3 of the dividend and consider 
this figure 3 as prefixed to the next figure 6 of the divi- 
dend, making 36. 

Now divide this number 36 by the divisor 12, saying 
12 into 36 goes 3 times. 

Place the quotient figure 3 obtained directly under 
the figure 6 of the dividend. 

The next figure 4 of the dividend is less than the 
divisor, in which case place directly under it a cypher (0) 
in the quotient and consider the figure 4 to be annexed 
to the next figure 8 of the dividend. 

Proceed now to divide the number 48 thus obtained 
by the divisor 12, saying 12 into 48 goes 4 times. 

Place the quotient figure 4 directly under the figure 8, 
and the division is completed. 

Examples for Practice 
5)402010 6)481236 '7)700245 



80402 


80206 


100035 


8)643424 


11)10124313155 


12)112862496 


80428 


920392105 


9405208 



When, after dividing the last figure of the dividend 
there is a remainder, the remainder is placed at the 
right of the quotient and underneath it set the divisor. 
The remainder and divisor when thus placed are sepa- 
rated by a short horizontal line ( — ). 



DIVISION 49 

Example. Divide 38 by 3. 

Arrange the divisor and dividend as before. 

3)38_ 

Divide the dividend figure 3 by the divisor 3, saying 
3 into 3 goes 1 time. 

Place the figure 1 obtained in the quotient. 

Then divide the next figure 8 of the dividend by the 
divisor 3, saying 3 into 8 goes 2 times and 2 over (re- 
mainder). 

Place the quotient figure 2 underneath the figure 8 
of the dividend, and place the remainder 2 with the 
divisor on the right of the quotient as shown. 

Example. Divide 4023 by 5. 

Arrange the numbers to be divided one by the other. 
Thus: 

5 )4023 

804| 

Proceed to divide, saying 5 into 40 goes 8 times. 
Place the 8 obtained in the quotient. 

Then say 5 into 2 goes times. 

Place the cypher (0) underneath the figure 2 of the 
dividend. 

Then say 5 into 23 goes 4 times and 3 over. 

Place the figure 4 obtained in the quotient underneath 
the figure 3 of the dividend. And the figure 3 which 
was left over, is a remainder, and is placed on the right 
of the quotient, having underneath it the divisor and 
between them a short horizontal line as shown in the 
example. 



50 A HAND BOOK FOR MECHANICS 

Examples for Practice 

5 )402013 6 )481237 7 )700243 8 )643226 
80402| 80206J 100034? 80403j 

11)10124313158 12)112862492 



920392105ft 9405207& 

To divide by any number up to 12, having a cypher (0), 
or any number of cyphers after it. 

That is, to divide by 2 with a cypher (0) or cyphers 
after it which would be either 20, 200, 2000, etc., or 5 
with a cypher or any number of cyphers after it, or 6 
with a cypher or any number of cyphers after it, as, 60, 
600, 6000, etc., up to 12 with a cypher or any number 
of cyphers after it. As 120, 1200, 12000, 120000, etc. 

Mark off from the right of the dividend as many 
figures as there are noughts (cyphers) in the divisor. 

Mark off also the noughts (cyphers) in the divisor. 

The remaining figures then of the dividend are divided 
by the remaining number of the divisor. 

Example. Divide 2469204 by 20. 

20 )2469204 
123460-4-#o 

Arrange the divisor 20 and dividend 2469204 in the 
proper manner, then mark off from the right of the 
dividend, 2469204, as many figures as there are cyphers 
in the divisor 20, in which case there will be one figure 
marked off, because there is one cypher in the divisor. 

Having done this, as shown in the examples, the 
remaining figures, 246920, of the dividend are divided 
by the remaining number 2 of the divisor, the cypher 
being marked off, by which operation the quotient, 
123460-4, is obtained. 



DIVISION 51 

The figure or figures which are marked off in the 
dividend are in all cases a remainder. 

Example. Divide 129876347 by 1200. 

120 0)129876347 

108230-347- xWo 

After arranging the numbers to be divided one by the 
other as before, mark off from the right of the dividend 
as many figures as there are cyphers in the divisor, 
which cyphers are also marked off. The dividend then 
becomes 1298763 and the divisor 12, and by dividing 
we obtain the quotient 108230, with a remainder of 3. 

In all cases where there is a remainder, place it in the 
quotient before the figures of the dividend, which are 
cut off. In which case this remainder 3 is placed before 
the figures 47 of the dividend which were cut off, making 
a remainder of 347, or j\Vo as shown. 

Examples for Exercise 



1. 


4023 by 5 


7. 


2045768 by 9 


2. 


168265 by 4 


8. 


204050 by 50 


3. 


8246 by 5 


9. 


364226734 by 2400 


4. 


523412 by 6 


10. 


92643266 by 11 


5. 


642139 by 7 


11. 


14465472 by 12 


6. 


904656 by 8 







When dividing by any number exceeding 12, the oper- 
ation is performed by a process called "Long Division," 
and when arranging the numbers to be divided one by 
the other, by the process of long division, set down the 
divisor and dividend as in short division, and on the 
right of the dividend place the quotient. 

The dividend and quotient are separated by a curved 



52 A HAND BOOK FOR MECHANICS 

line similar to the curved line placed between the divisor 

and dividend. 

Hence the numbers in long division are arranged as 

follows: 

Divisor) Dividend (Quotient 

which shows the quotient placed on the right of the 
dividend. 

When dividing a number which is greater than 12 
into another number, inquire the number of times the 
divisor is contained in the least number of figures on 
the left of the dividend. 

Place the figure obtained by the operation in the 
quotient. 

Multiply the figures of the divisor by the quotient 
figure and place the product under those figures of the 
dividend in which the divisor operated. 

Example. Divide 26 by 13. 

Arrange the divisor, dividend, and quotient places as 

follows: 

Dividend 

Divisor 13)26(2 Quotient 

26 

00 

Proceed then to divide the dividend 26 by the divisoi 
13. saying 13 into 26 goes 2 times. 

Place the figure 2 obtained, in the quotient, which is 
on the right of the dividend. 

Then multiply the figures 13 of the divisor by the 
quotient figure 2. First multiply the figure 3 of the 
divisor by the quotient figure 2. saying 2 times 3 are 6, 
which place underneath the figure 6 of the dividend. 

Then multiply the next figure 1 of the divisor by the 



DIVISION 53 

quotient figure 2, saying 2 times 1 are 2, which place 
underneath the figure 2 of the dividend. 

Draw underneath the product 26 obtained by the 
multiplication, a horizontal line thus ( — ) 

Subtract the product 26 from the dividend 26 and the 
division is completed. Showing that 13 is contained in 
26 2 times. 

Example. Divide 75 by 15. 

Arrange the divisor, dividend, and quotient as before. 

15)75(5 

75 

00 

Proceed then to divide the dividend 75 by the divisor 
15, saying 15 into 75 goes 5 times. Place the figure 5 
in the quotient. 

Then multiply the two figures 15 of the divisor by 
the quotient figure 5, saying 5 times 5 are 25. 

Place the right-hand figure 5 of this product 25 under 
the figure 5 of the dividend and add the left-hand figure 
2 of the product to the product of the figure 1 when 
multiplied by the quotient figure 5, saying 5 times 1 
are 5 and 2 are 7. 

Place the product figure 7 under the dividend figure 7. 

We have now obtained the product 75, which is placed, 
as explained, under the figures 75 of the dividend. 

Proceed then to subtract, saying 5 from 5 leaves 0. 
Then 7 from 7 leaves 0, and the division is completed. 

Examples for Exercise 



14)56(4 


13)39(3 


15)60(4 


16)80(5 


56 


39 


60 


80 


00 


00 


00 


00 



54 A HAND BOOK FOR MECHANICS 



17)34(2 
34 


18)72(4 
72 


00 


00 



Process. When there is a remainder after subtracting 
the product, which is obtained by multiplying the divisor 
by the quotient figure, place the remainder in the quo- 
tient and underneath it a short horizontal line, under- 
neath which place the divisor, as in short division. 

Example. Divide 53 by 13. 

Arrange the numbers as before. 

13)53(4x3 
52 

1 

Then divide the dividend 53 by the divisor 13, saying 
13 is contained in 53 4 times. 

Place the figure 4 obtained in the quotient. 

Now multiply the divisor 13 by the quotient figure 4, 
placing the product obtained by the operation under- 
neath the figure 53 of the dividend in which the divisor 
13 operated. 

Proceed now by subtracting the product 52 from the 
dividend 53, saying 2 from 3 leaves 1 (remainder) and 
5 from 5 leaves nothing. Place the figure 1 (remainder) 
in the quotient, with the divisor underneath it as shown 
in the example. 

Example. Divide 69 by 14. 

Arrange the numbers as before. 

14)69(411 
56 
13 



DIVISION 55 

Inquire the number of times 14, the divisor, is con- 
tained in 69, the dividend, saying 14 into 69 goes 4 
times. 

Place the 4 obtained in the quotient and use it as a 
multiplier, with the divisor 14 as the multiplicand, by 
which operation 56 is obtained, which is placed, as shown, 
underneath the figure of the dividend in which the 
divisor 14 operated. 

Proceed now to subtract, saying 6 from 9 leaves 3 
and 5 from 6 leaves 1. 

By the subtraction we have 13 (remainder), which 
place in the quotient with the divisor 14 underneath it, as 
shown, and the division is finished. 

Examples for Practice 

14)38(211 14)58(4t 2 t 26)94(3M 572)983(ltH 
28 56 78 572 

10 2 16 ill 

Process. When there are more figures in the divi- 
dend than there are in the product to be subtracted 
from it, we place on the right of the number obtained 
by the subtraction the figure next to the figure or group 
of figures in which the divisor operated. The number 
thus formed is the dividend in which the divisor operates. 

Example. Divide 9424 by 38. 

Arrange the divisor and dividend as before. 

38)9424(248 
76 

182 
152 

304 
304 



56 A HAND BOOK FOR MECHANICS 

Then inquire the number of times the divisor 38 is con- 
tained in the 94, the two left-hand figures of the dividend, 
and finding it is 2 times, place the 2 in the quotient. 

Now multiply the divisor 38 by the quotient figure 2, 
and subtract the product 76 obtained by the multiplica- 
tion from dividend figures 94, and to the right of the 
remainder 18 bring down the next figure 2 of the dividend, 
making 182. 

The number 182 thus formed is the dividend in which 
the divisor now operates. 

Proceed, inquiring how many times 38 is contained 
in 138, finding it to be 4 times. 

The 4 is placed in the quotient. Now multiply the 
divisor 38 by 4 and subtract the product 152, which is 
obtained, from 182, and to the right of the remainder 
30 bring down the next figure 4 of the dividend. The 
number 304 thus formed is the dividend in which the 
divisor next operates, in which case proceed, saying 38 
is contained in 304 8 times. 

Place the 8 in the quotient and multiply the divisor 
38 by it and subtract, as before, the product obtained 
from the dividend [304]. 

In which case the division is finished. 

Example. Divide 14894 by 52. 

Arrange the divisor, dividend, and quotient as before. 

52)14894(28611 
104 

449 
416 

334 
312 

~22 



DIVISION 57 

Proceed to divide, inquiring how many times the 
divisor 52 is contained in the least number of figures in 
the left of the dividend. 

It is found that 52, the divisor, is contained in 148 
2 times. 

Place the 2 in the quotient, multiply the divisor by it, 
and place the product, 104, obtained by the operation 
underneath the figures 148 of the dividend. 

Subtract the product 104 from the dividend figures 
148; by so doing we get a remainder of 44, on the right 
of which place the figure 9 of the dividend. 

The number 449 thus formed is the dividend in which 
the divisor now operates. 

Proceed, inquiring the number of times the divisor 52 
is contained in 449, and finding it to be 8 times, place 
the 8 in the quotient, multiply the divisor 52 by it and 
subtract the product 416, obtained -by the operation, 
from the dividend 449; by so doing we have a remainder 
of 33, on the right of which we place the next figure 4 
of the dividend. The number 334 thus formed is now 
divided. 

In 334, 52 is contained 6 times. Place the 6 in the 
quotient, multiply the divisor 52 by it, and subtract the 
product 312, obtained by the operation, from the dividend 
334. 

By the subtraction there is a remainder of 22, which 
place in the quotient with the divisor 52 underneath it, 
as shown. 



58 A HAND BOOK FOR MECHANICS 

Examples for Practice 



67)89726(1339-^ 


223)6994234(31364^ 


67 


669 


227 


304 


201 


223 


262 


812 


201 


669 


616 


1433 


603 


1338 


13 


954 




892 




62 


25)7284600(291384 


7645)72698 l(95#ft 


50 


68805 


228 


38931 


225 


38225 


34 


706 


25 




96 




75 




210 




200 




100 




100 





Process. When there are less figures in the product 
obtained by multiplying the divisor by a quotient figure 
than there are in the group of figures in which the divisor 



DIVISION 59 

operates, in all cases place the right-hand figure of the 
product directly underneath the right-hand figure of 
the group of figures in which the divisor operates. 
For instance, divide 1256 by 85. 

Arrange the divisor 85, dividend 1256, and quotient as 
before. 

85)1256(1411 
85 

406 
340 



66 



It is seen here that the two figures 12 on the right of 
the dividend will not contain the divisor 85, so we inquire 
the number of times the divisor 85 is contained in the 
group of three figures on the left of the dividend. 

Finding that it is contained in 125 1 time, place the 1 
in the quotient, multiply the divisor by it and set the 
product underneath that group of figures, 125, in which 
the divisor 85 operates, remembering to place the right- 
hand figure 5 of the product 85 underneath the right-hand 
figure 5 of the group of figures in which the divisor 
operates. 

Now subtract the product 85 from the dividend 125, 
by which operation we get 40; bring down the figure 6 
of the dividend forming the number 406, into which the 
divisor goes 4 times. 

Place the 4 in the quotient, multiply the divisor by it, 
and subtract the product 340, obtained by the opera- 
tion, from the dividend 406, by doing which we get a 
remainder of 66, which place as shown in the quotient, 
and the division is completed. 

Example. Divide 20678 by 98. 



60 A HAND BOOK FOR MECHANICS 

98)20678(211 
196 



107 

98 



98 
98 



Having arranged the divisor, dividend, and quotient 
places as before, inquire the number of times 98, the 
divisor, is contained in 206, the three left-hand figures 
of the dividend, and finding it to be 2 times, we place 
the 2 in the quotient, multiply the divisor 98 by it and 
subtract the product (196), obtained by the multiplica- 
cation, from the group of figures 206, and to the right 
of the remainder 10, obtained by the subtraction, place 7, 
which is the figure next of the dividend to the group of 
figures in which the divisor operates, making 107. 

We next inquire the number of times 98 is contained 
in 107 and, finding it to be 1 time, place the 1 in the 
quotient, then multiply and subtract as before. 

Note that the product 98 contains less figures than 
the group of figures which is divided, in which case, as 
before stated, the right-hand figure of the product is 
placed underneath the right-hand figure of the group of 
figures which form the dividend, so the 8 of the product 
is placed under the 7 of the dividend. By subtracting 
the product 98 from 107 there is a remainder of 9, 
on the right of which place the figure 8, the last figure 
of the dividend 20678, and proceed as before, saying 98. 
the divisor, goes into 98, the dividend, in which it 
operates 1 time. Place the 1 in the quotient and mul- 
tiply the divisor 98 by it. After finding the product, 
no remainder is left. 



Examples 


FOR 


Practice 


89) 46386 (52 lH 




596)131040(21911 


445 




1192 


188 


1184 


178 




596 


106 




5880 


89 




5364 


17 




. 516 



61 



It often happens, when a figure is taken from the 
dividend and placed on the right of the partial divi- 
dend, the number thus formed will not contain the 
divisor. 

When such is the case, place a cypher in the quotient 
and bring down another figure from the dividend, which 
place on the right of the partial dividend, and so on 
till the partial dividend is large enough to contain the 
divisor. 

Example. Divide 7828 by 76. 

76)7828(103 
76 



228 
228 



Having arranged the divisor, dividend, and quotient 
places, inquire how many times 76 is contained in the 
figures 78 on the left of the dividend and, finding it to 
be 1 time, place the 1 in the quotient, multiply the 
divisor 76 by it, and subtract the product 76, obtained 
by the operation, from the 78, and to the right of the 
remainder 2 place 2, the next figure of the dividend, 



62 A HAND BOOK FOR MECHANICS 

making 22. It is seen that 22 will not contain the 
divisor 76, so we place a cypher (0) in the quotient, and 
place on the right of the 22 the next figure 8 of the 
dividend. By so doing we have a partial dividend, 
228. 

Inquire now how many times the divisor 76 is con- 
tained in the number 228 thus formed and, finding it to 
be 3 times, place the 3 in the quotient, multiply the 
divisor 76 by it, and subtract the product 228, obtained 
by the operation, from the partial dividend 228. By 
the subtraction there is no remainder. 

Example. Divide 1197965 by 597. 

597)1 197965 (2006 W 
1194 



3965 

3582 
383 

Inquire the number of times the divisor 597 is con- 
tained in the least number of figures on the left of the 
dividend and, finding it to be contained in the figures 
1197 2 times, place the 2 in the quotient, multiply the 
divisor 597 by it; subtract the product, 1194, obtained 
from the dividend figures 1197. By the subtraction we 
have 3 left, on the right of which place the next figure 9 
of the dividend. 

The number 39 thus formed will not contain the 
divisor, so we place a cypher (0) in the quotient, and 
place on the right of the 39 the next figure 6 of the 
dividend, making 396. Again we have a number which 
will not contain the divisor, so again place a cypher (0) 
in the quotient. 

Then bring down the next figure 5 of the dividend, 



TABLES 63 

placing it on the right of the figures 396, making 3965 
as a partial dividend. Then the divisor 597 is seen to 
be contained in 3965, the number thus formed, 6 times. 
Place the 6 in the quotient, multiply the divisor by it, 
and subtract the product from the partial dividend 3965. 



Divide: 



Examples for Exercise 

1. 76298764833 by 9. 

2. 120047629817 by 20. 

3. 9876543216 by 48. 

4. 3247617219 by 63. 

5. 7141072617 by 41. 

6. 329817228 by 107. 

7. 24769876237 by 987. 

8. 2610014723 by 2406. 

It will be noticed that three arithmetical rules are 
required to perform operations in long division, namely: 
division, multiplication, and subtraction. 

TABLES 

(Tables of weights and measures used by mechanics.) 

Capacity 

4 gills make 1 pint 4 pecks make 1 bushel. 

2 pints make 1 quart. 8 bushels make 1 quarter. 

4 quarts make 1 gallon. 5 quarters make 1 load. 
2 gallons make 1 peck. 

The long ton of 2240 avoirdupois pounds is used at 
"custom houses, and in weighing coal at the mines. 



64 A HAND BOOK FOR MECHANICS 

Length 

12 inches make 1 foot. 
3 feet make 1 yard. 
6 feet make 1 fathom. 
h\ yards make 1 rood, pole, or perch. 
40 poles make 1 furlong. 
8 furlongs make 1 mile. 
3 miles make 1 league. 
5280 feet or 1760 yards make 1 geographical mile. 
6080 feet make 1 nautical mile. 

Surface 

144 square inches make 1 square foot. 

9 square feet make 1 square yard. 
30 \ square yards make 1 square rood, pole, or perch 
40 perches make 1 rood. 
4 roods make 1 acre. 
640 acres make 1 square mile. 

Solidity 

1728 cubic inches make 1 cubic foot. 
27 cubic feet make 1 cubic yard. 

Dimensions 

12 inches make 1 foot. 
144 square inches make 1 square foot. 
1728 cubic inches make 1 cubic foot. 
5280 feet make 1 geographical mile 
6080 feet make 1 nautical mile or knot. 

Time 

60 seconds (sec.) make 1 minute (mi.). 
60 minutes make 1 hour (h.). 



TABLES 65 

24 hours make 1 day (d.). 

365 days make 1 common year (y.). 

366 days make 1 leap year (1. y.). 
7 days make 1 week (wk.). 

12 calendar months (mo.) make 1 year. 
100 years make 1 century. 

The " Civil Day" begins at 12 o'clock midnight. 

Weights 

Troy 

24 grains (gr.) are 1 pennyweight. 
20 pennyweight are 1 ounce (oz.). 
12 ounces are 1 pound (lb.). 

Avoirdupois 

16 ounces (oz.) are 1 pound (lb.). 
100 pounds are 1 cental (ctl.). 
2000 pounds are 1 ton (T.). 

A pound avoirdupois is equal to 7000 grains, and a 
pound troy is equal to 5760 grains, so that 144 pounds 
avoirdupois are equal to 175 pounds troy. 

Circular Measure 

60 seconds (") are 1 minute ('). 
60 minutes are 1 degree (°). 
360 degrees are 1 circumference (cir.). 

The circumference of a circle is supposed to be divided 
into 360 equal parts, called degrees. 

A degree is therefore ih of the circumference of a 
circle. 



66 A HAND BOOK FOR MECHANICS 

A quadrant is a fourth of a circumference, or an arc 
of 90 degrees. 

A degree is divided into 60 minutes expressed by 
sign (') and each minute is divided into 60 seconds 
expressed by sign ("), so that the circumference of any 
circle contains 21,600 minutes, 1,296,000 seconds. 

Wood Measure 

16 cubic feet are 1 cord foot (cd. ft.). 
8 cord feet or 128 cubic feet are 1 cord (cd.). 

Weights of Water 

1 gallon of fresh water weighs 10 pounds. 

1 gallon of sea water (specific gravity of 1029) weighs 

10j pounds. 
1 gallon equals tVo (.16) of a cubic foot, or 1 cubic foot 

equals 6i gallons. 
1 cubic foot of fresh water weighs 62| pounds, equal to 

1000 ounces. 
1 cubic foot of sea water weighs 64 pounds. 
35 cubic feet of sea water weigh 1 ton. 
35.84 (35t 8 o 4 o) cubic feet of fresh water weigh 1 ton. 

Weights of Metals 

Wrought iron 3.6 cubic inches weigh 1 pound; or 1 cubic 

inch equals .2778 of a pound. 
Cast iron 3.9 cubic inches weigh 1 pound; or 1 cubic 

inch equals .257 of a pound. 
Soft steel 3.55 cubic inches weigh 1 pound; or 1 cubic 

inch equals .2814 of a pound. 
Brass or gun metal 3.3 cubic inches weigh 1 pound; or 

1 cubic inch equals .3 of a pound. 



REDUCTION 67 



REDUCTION 

Reduction is the process of changing numbers from 
one denomination to another denomination of the same 
general kind without altering their values. 

By denomination is meant name. 

Thus 36 feet is a denomination. 

Also in 3 days, 7 hours, 50 minutes, there are three 
denominations (days, hours, and minutes). 

It is possible by the process of reduction to change 
number of a higher denomination to a lower denomina- 
tion without altering their value. 

For example: 3 feet may be reduced to 36 inches. 

Here feet is a higher denomination than inches, because 
it takes a certain number of inches to make a foot, But 
by reducing 3 feet to 36 inches the value is not changed, 
as 3 feet and 36 inches represent the same length. 

It is also possible by the process of reduction to change 
numbers of a lower denomination to numbers of a higher 
denomination and not alter their values. 

Inches may be reduced to feet, for instance. 

Thus 36 inches reduced to feet is 3 feet. 

Here inches, the lower denomination, is reduced to 
feet, the higher, but the values are unchanged, as 36 
inches and 3 feet, as stated before, are equal. 

Process. When reducing a higher to lower denomi- 
nations, multiply the higher denomination which is to 
be reduced by the number of units it takes of the next 
lower denomination to equal one of that higher denom- 
ination, and then to the product add the given numbers, 
if any, of the smaller denomination. 

Example. Reduce 27 pounds to ounces. 

As the next lower denomination to pounds is ounces, 



68 A HAND BOOK FOR MECHANICS 

we will multiply the 27 pounds by the number of units 
it takes of that denomination (ounces) to make 1 pound. 
Then, as 16 ounces make 1 pound, 27 pounds mul- 
tiplied by 16 ounces will give us the number of ounces 
contained in 27 pounds. Thus: 

27 
16 



162 

27 



432 oz. Ans. 

Example. How many ounces are there in 25 lb. 7 oz.? 
As 16 ounces make a pound, multiply the 25 pounds 
by 16 as follows: 

25 lb. 7 oz. 
16 



150 
25 

400 

7 



407 oz. 

The product of which is 400 ounces, and to the product 
add the 7 ounces, making 407 ounces. 

Hence in 25 lb. and 7 oz. there are 407 ounces. 

Example. How many grains are there in 4 penny- 
weight (dwt.) troy? 

Now 24 grains make a pennyweight, therefore multiply 
the 4 by 24 as follows: 

4 
24 

96 gr. Ans. 



REDUCTION 69 

Example. How many pennyweights are there in 5 
ounces (troy) ? 

20 pennyweights make an ounce, therefore to find 
the number of pennyweights in 5 ounces, multiply the 
5 by 20 as follows: 

5 
20 

100 dwt. Ans. 

Example. How many ounces are there in 16 pounds 

(troy) ? 

Now 12 ounces make a pound (troy), therefore multi- 
ply the 16 by 12 as shown. 

16 1b. 
12 oz. 

192 oz. Ans. 

Example. How many grains are there in 32 lb. 
10 oz. 15 dwt. 7 gr. (troy) ? 

32 lb. 10 oz. 15 dwt. 7 gr. 
12 

394 oz. 
20 



7880 
15 



7895 dwt. 
24 



31580 
15790 



189487 gr. Ans. 

Bring the 32 pounds to ounces by multiplying by 12 
(12 ounces equals a pound troy). 



70 A IIANIj HOOK FOR MECHANICS 

To the product 384 add the 10 ounces, making 394 
ounces. 

Now bring the 394 ounces to pennyweights by mul- 
tiplying by 20 (there being 20 pennyweights in one 
ounce). 

To the product 7880 add the 15 pennyweights, making 
7895 pennyweights. 

Proceed now to convert the 7895 pennyweights into 
grains. 

A grain equals 24 pennyweights, therefore multiply 
7895 by 24 and to the product add the 7 grains. 

And, as shown, in 32 lb. 10 oz. 15 dwt. 7 gr. there are 
189487 grains. 

Example. Reduce 3 yd. 2 ft. 9 in. to inches. 

Bring the yards to feet, then bring the feet to inches 
as follows: 

3 yd. 2 ft. 9 in. 
3 

11 ft. 
12 



141 in. Arts. 

As shown, we bring the 3 yards to feet by multiplying 
by 3, there being 3 feet in one yard, add 2 feet to the 
product, making 1 1 feet. We then bring the 1 1 feet to 
inches by multiplying by 12, because there are 12 inches 
in one foot, and to the product add the 9 inches, and, as 
shown, in 3 yd. 2 ft. 9 in. there are 141 inches. 

Process. When reducing lower to higher denomina- 
tions, divide first the given number by the number of 
units it takes of that denomination to equal one of the 
next larger, and reserve the remainder, if any, and 
divide next the quotient and so proceed until the re- 



REDUCTION 71 

quired denomination is reached. The last quotient and 
the several remainders, if any, will be the number sought. 

Example. How many pounds are there in 407 ounces? 
In other words, reduce 407 ounces to pounds. 

Divide first the given number, 407 ounces, by that 
number of units it takes of that denomination of ounces 
to make one of the next larger denomination. The next 
larger denomination is pounds, and as it takes 16 ounces 
to make one pound, we divide 407 by 16. Thus: 

16)407 
25-7 

Dividing, then, the 407 ounces by 16 reduces it to 25 
pounds, and the remainder 7 is ounces. 
The above answer then reads 25 lb. 7 oz. 

Example. How many pounds are there in 179 ounces? 

Here, as before, to bring the 179 ounces to pounds, 
divide by 16, because there are 16 ounces in a pound 
avoirdupois, and if there is a remainder it is called 
ounces, as follows: 

16)179 

11-3 Ans. 11 lb. 3 oz. 

Example. How many yards in 141 inches? 

First bring the 141 inches to feet by dividing by 12, as 
follows: 

12 )141 
3)11-9 in. 
3-2-9 

By so doing we have a quotient of 11 and 9 over, 
which are inches. Next bring the 11 feet to yards by 
dividing by 3, because 3 feet is one yard, as shown. And 
we obtain the quotient of 3 yards 2 feet. 



72 A HAXD BOOK FOR MECHANICS 

Hence in full the answer is 3 ft. 2 yd. 9 in. 

Example. How many pounds are there in 189487 
grains? 

24 )1894 87 
2.0 ) 7895-7 gr. 
12) 394.15 dwt. 
32.10 oz. 
Ans. 32 lb. 10 oz. 15 dwt. 7 gr. 

FRACTIONS 

The term fraction is derived from the Latin word 
frango, which means to break or separate into parts. 

A fraction, then, means a part of a whole thing or a 
part of any number. And by the use of fractions we are 
enabled to perform arithmetical operations where numbers 
less than whole numbers are employed. There are two 
kinds of fractions, called vulgar fractions and decimal 
fractions, each of which are represented differently. This 
chapter pertains to vulgar or common fractions only. 

A vulgar or common fraction is always represented 
by not less than two numbers, one of which is placed 
over the other, having a short horizontal or diagonal line 
between them. Thus: 

t or H 

A horizontal line is — and a diagonal line is / . 

That number which is above the line is called the 
numerator, and the number underneath the line is called 
the denominator. Thus in the above fraction 3 is the 
numerator and 8 is the denominator. The denominator 
always denotes the number of equal parts there are in a 
whole thing. The numerator always denotes the num- 



FRACTIONS 



73 



ber of those equal parts we have. Thus in the fraction 
f, the 8 tells us the whole thing has been divided into 
eight equal parts, and the 3 tells us that there are three 
of those equal parts represented. 

For instance, take some specific length, say one inch, 
and divide it into a number of equal lengths, say four. 
Thus: 



2 3 4 



-l-NC- 



If it is required now to specify in figures a number 
that will represent a length equal to one of the divisions, 
we would represent it thus, J inches. The denominator 
4 would show that the inch (the whole thing) is divided 
into four equal parts, and the numerator would tell the 
number of those parts represented. Likewise, } inches 
would represent a dimension equal in length to three 
of the four equal divisions into which the whole inch 
is divided. Practically speaking, vulgar fractions are 
divided into six different classes, named: 

1. Proper fractions. 4. Compound fractions. 

2. Improper fractions. 5. Simple fractions. 

3. Mixed fractions. 6. Complex fractions. 

First. A Proper fraction is a fraction whose numerator 
is less than its denominator. 



As | 



or h or 



Second. An Improper fraction is a fraction whose 
numerator is equal to or greater than its denominator. 

As § or § or | or | 



74 A HAND BOOK FOR MECHANICS 

Third. A Mixed number is a whole number with a 
fraction. 

As 8$ or 5| or 6i 

Fourth. A Compound fraction is a fraction of a 
fraction. 

As | of }, or f of f , or J of \ 

Fifth. A Simple fraction has but one numerator and 
one denominator, and may be either improper or proper. 

As | or | or | or f 

Sixth. A Complex fraction is a fraction which has 
either a compound fraction, a mixed number, a whole 
number, or a simple fraction, for either its numerator or 
denominator, or both. 

3 

As -— > 

7 

8 

an example of a complex fraction having simple 
fractions for numerator and denominator; 

6 
or — * 
3| 

example of complex fraction having whole number 
for numerator and mixed number for denominator; 



or 



51 



example of complex fraction having mixed numbers 
for numerator and denominator. 

REDUCTION OF FRACTIONS 

Reduction of fractions is the process of changing frac- 
tional terms without altering their values. The "terms" 
of a fraction means its numerator and denominator. 



REDUCTION OF FRACTIONS 75 

To reduce a fraction to its lowest terms, divide the 
numerator and denominator by any number that will 
divide them both without a remainder. 

Example. Reduce t 8 2 to its lowest terms. Thus: 

4)_§_ 2 
12 3 

Here 4 will divide both numerator 8 and denominator 
12 without a remainder. 4 goes into 8 two times and 
into 12 three times. Hence f is equal to t 8 2 In other 
words, ^ of an inch is the same part of an inch (or 
length) as § of an inch is, or § of an apple would represent 
a piece of an apple the same size as t 8 2 of the same apple. 
Therefore, by bringing the fraction T 8 2 to its lowest terms, 
f, the value of the fraction, has not been altered. The 
terms of the fraction were changed only. 

Reduction of fractions, then, is performed by dividing 
the numerator and denominator by any number that 
will divide them both without a remainder, and the 
successive results of the division are divided by any 
number that will divide them both without a remainder, 
until the operation of dividing can be carried no further. 

Example. Reduce ft to its lowest terms. 

8)24 3 , 
72 9 3 

Here the number 8 will divide the numerator and 
denominator without a remainder, and by the division 
is obtained f . We next divide this result by 3, because 
3 will divide both 3 and 9 without any remainder, and 
obtain J, which cannot be further divided; therefore, J 
is the fraction f f reduced to its lowest terms, and both 
fractions represent the same value. 



76 A HAND BOOK FOR MECHANICS 

Example. Reduce iWA to its lowest terms. 

First divide the numerator and denominator by 12, 
and by the division the terms become 1W2. Then divide 
again by 12 and the terms tW* become M. 12 will 
again divide them and it becomes J, which is the lowest 
term. Consequently, J of an inch is the same length as 
tVsV? of an inch. 

Examples for Exercise 

Reduce t% to its lowest terms. 
Reduce to to its lowest terms. 
Reduce it to its lowest terms. 
Reduce if to its lowest terms. 
Reduce if to its lowest terms. 
Reduce tWs to its lowest terms. 
Reduce Iff to its lowest terms. 

To reduce a fraction of any denominator to a fraction 
having a greater denominator, that is, to bring a fraction 
to its largest terms, which is the reverse of the last rule, 
see how many times the denominator of the fraction in 
sight will divide into the larger denominator to which 
you are to bring it. 

Example. Bring J to a fraction whose denominator 
is 8. 

Proceed by dividing the denominator 2 of the fraction 
\ into the greater denominator 8, and multiply the 
numerator and denominator of the fraction \ in sight 
by the quotient obtained by the division. Here 2 goes 
into 8 four times. Then multiply the numerator and 
denominator of the fraction \ by the quotient 4, and by 
the multiplication is obtained |. 



Arts. 


i 


Arts. 


i 


Arts. 


§ 


Arts. 


j 


Arts. 


i 
3- 


Arts. 


i 


Ans. 


236 
4T7 



REDUCTION OF FRACTIONS 77 

Multiplying both terms of a fraction by the same 
number does not change its value, hence, 

1 _ 4 

2 — 8- 

That is, J of an inch would be equal to | of an inch. 

Example. Bring § to a fraction whose denominator 
is 15. 

Here 3 goes into 15 five times, then § becomes if- 

Examples for Exercise 

Bring f to a fraction whose denominator is 12. Arts. tV 
Bring J to a fraction whose denominator is 16. Ans. if • 
Bring f to a fraction whose denominator is 20. Ans. |§. 
Bring J to a fraction whose denominator is 8. Ans. f . 

When bringing a fraction to its largest terms, as in 
the above examples, the denominator of the fraction 
whose terms are to be changed must divide into the 
denominator to which it is to be changed without a 
remainder. 

To reduce a mixed number to an improper fraction, 
multiply the whole number by the denominator of the 
fraction, and to the product add the numerator, and 
place their sum over the denominator of the fraction. 

Example. Reduce 6 J to an improper fraction. 

Multiply the whole number 6 by the denominator 8 of 
the fraction, and add to the product obtained by the 
multiplication the numerator 7, saying 8 times 6 are 48 
and 7 are 55. Place the sum 55 thus obtained over the 
denominator 8 of the fraction, as follows: 

55 

Then 5 gf represents the mixed number 6| when reduced 



Arts. 


1 9 
4 


Ans. 


47 

s • 


Ans. 


20 
3 • 


Ans. 


53 
5 • 


Ans. 


406 



78 A HAND BOOK FOR MECHANICS 

to an improper fraction, and it shows that there are 
contained in 6J, 55 eighths. That is, there would be 
contained in 6J inches 55 eighths of an inch. 

For instance, in one inch there are 8 eights, therefore in 
6 inches there are just six times as many eighths, which 
would be 48 eighths, to which add seven more eighths 
and you have 55 eighths. 

Examples for Exercise 

Reduce 4§ to an improper fraction. 
Reduce 5 J to an improper fraction. 
Reduce 6§ to an improper fraction. 
Reduce 101 to an improper fraction. 
Reduce 67f to an improper fraction. 

To reduce an improper fraction to a whole or mixed 
number, divide the numerator by the denominator, and 
if there is a remainder, place it over the denominator at 
the right of the whole number obtained by the division. 

Example. Change ff to a mixed number. 

Divide the numerator 24 by the denominator 16 as 
follows, and place the remainder ^ on the right of the 
whole number 1, obtained by the division. 

16)24(11 

^6 

1=1 

n 2 

Examples for Exercise 

Reduce - 1 ? 2 - to a whole number. Ans. 3. 

Change V - to a mixed number. Ans. 14$. 

Reduce *£■ to a whole number. Ans. 15. 

Change Vi 3 " to a mixed number. Ans. 48t4- 



REDUCTION OF FRACTIONS 79 

To reduce a compound fraction to a single fraction: 
Example. Reduce J of J to a single fraction. 

Multiply all the numerators together for the new 
numerator (that is, for the numerator of the single frac- 
tion), and all the denominators together for the new 
denominator; then, if necessary, reduce the fraction 
obtained to its lowest terms. Bear in mind that the 
sign X is read " multiplied by" and means that the num- 
ber before it is multiplied by the number after it. 

Proceed now to reduce the above compound fraction 
to a simple, or single, fraction, as follows: 

I X i = U- Ans. 

As shown, the numerator 3 and 7 are multiplied to- 
gether for the new numerator, which is 21, and the 
denominators 4 and 8 are multiplied together for the 
new denominator, which is 32. We then have the 
fraction %\. 

Example. Reduce § of J of T V to a single fraction. 

As before, multiply all the numerators together for 
the new numerator, and all the denominators for the 
new denominator, and, if necessary, reduce the fraction 
obtained by the multiplication to its lowest terms, as 
follows: 2 17 U J_ . 

3 X 5 X 16~pFl20 AnS - 

By multiplying all numerators together we get a new 
numerator, 14, and by multiplying the denominators 
together we get a new denominator, 240, both of which 
may be divided by 2 without a remainder, as shown. 
By a process of canceling we can also reduce a compound 
fraction to a single fraction. 



80 A HAND BOOK FOR MECHANICS 

Rules for Canceling: 

Any numerator can be divided into any denominator, 
provided no remainder is left, and any denominator can 
be divided into any numerator, provided no remainder 
is left. 

Example. Reduce | of § of | to a single fraction. 
Thus: 8 ,2 ,4 

Here the 3 in the first numerator and the 3 of the 
second denominator will cancel one into the other; these 
are therefore left out. Also the 4 of the first denomi- 
nator cancels with the 4 of the last numerator, and then 
we have 2 left as a numerator and 9 as a denominator. Thus : 

2 

Many problems may be considerably shortened by the 
process of cancellation, as the following example will 
show. 

Example. Reduce § of f of \% of fft to a single 
fraction. Thus: 

? of ^ of M of ii = Z _JL 

P £ # m 3 X 2 X 34 204 
3 £ 34 

2 

The process is as follows: The first numerator, 2, will 
go into 8, the denominator of the second fraction, 4 times; 
the denominator 18, of the third fraction, will go into the 
90, the numerator of the last quantity, 5 times. The 
numerator 3 of the second fraction will go into the de- 
nominator 9 of the first fraction, 3 times. 5 will go 
into the denominator 170, of the last quantity, 34 times; 
2 will go into 4 twice, and into the numerator 14 of the 



REDUCTION OF FRACTIONS 81 

third fraction, 7 times. And as we cannot find any more 
figures that can be divided without leaving a remainder, 
we are at an end and the quantity left must be put into 
an expression. On examination, we have 7 left in the 
top row; this is put down at the end as the final numer- 
ator. On the bottom we have 3, 2, and 34; these multi- 
plied together give us 204, which is the final denominator. 
Any numerator and denominator may be divided by 
the same number, provided there be no remainder left, 
and the values obtained by the division are used in the 
place of the values that were canceled. 

Examples. Reduce f of §r to a single fraction. 

Here 8, the denominator of the first fraction, can be 
divided by 4, and 20, the numerator of the last quantity, 
can be divided by the same number without leaving any 
remainder, as shown. Thus: 

5 

3 ,?|_ 3X5 15 

JB 31 ~ 2 X 31 "" 62 Am ' 

2 

Example. Reduce -ft- of A of tt to a single fraction 
Thus: 

1 f A f M 7 _7_ 

J0 ffi 17 " 3 X 2 X 17 " 102 
3 i 
2 

Examples for Exercise 
Reduce to their simplest forms the fractions: 

i of § of | of J 
This can be done by canceling as follows. Thus: 



82 ^1 HAND BOOK FOR MECHANICS 

12-341 

7 of -z of 7 of - = - Arts. 

'1 p 4 5 o 

Reduce f of H of | to a single fraction. 

3 

3 ,16 4_ _3_ _3^ 

5 16 ° 9 ~ 4 X 3 ~ 12 
4 3 

Reduce f of J of }* to a single fraction. Thus: 

2 

p , 1 , 1 "4 1 1 

of 7 of — = 



/7 £ ^ 2x3 6 
2 3 

Reducing Whole Numbers to Improper Fractions 

Placing 1 for a denominator under a whole number 
reduces the whole number to an improper fraction. 

Example. Reduce 7 to an improper fraction. 

Place 1 under the whole number 7 and we have an 
improper fraction, whose numerator is 7 and whose de- 
nominator is 1. Thus: 



All whole numbers in a compound fraction must be 
converted into an improper fraction before multiplying 
the numerators together and the denominators together. 

Example. What is the value of f of f of 7. 

Notice that the last quantity,. 7, is a whole number. 
It is therefore converted into an improper fraction by 
placing 1 underneath it for a denominator, as follows: 

§ 4 7 = _ 4X7 = 28 
5 9 1 5 X 3 X 1 == lo)28(l T f 
3 15 

13 



REDUCTION OF FRACTIONS 83 

By first canceling and then multiplying the remaining 
numerators together for a new numerator, and then 
multiplying the remaining denominators together for the 
new denominator, we obtain the improper fraction ft, 
which is reduced to a mixed number. Then, as shown, 
the value of § of f of 7 = lit • 

All mixed numbers in a compound fraction must be 
reduced to an improper fraction before multiplying the 
numerators together and the denominators together. 

Example. Find the value of | of T 4 T of H of £§ of 5|. 

Notice the last value, 5f, is a mixed number, which, 
when reduced to an improper fraction is t 3 -. 

Then we have f of t 4 t of {$ of J§ of -£-. 

After canceling and multiplying together the remaining 
numerators for the new numerator, and the remaining 
denominators for the new denominator, we have, as 
shown: 

5 n 1% w i 5 

Examples for Exercise 

Reduce f of to of if to single fraction. Ans. #5. 
Reduce J of tt of 7 to mixed numbers. Ans. 4J|. 
What is § of tt of f of f. Ans. tIt- 
Change to a single fraction H of J of f of 2V of 8. 
Ans. 3T 3 o. 

Find the value of § of J of f of tt of li of 5|. Ans. f |. 

What is I of 2|. Ans. 2 T \. 

Reduce tr of ft of if of 9f to a whole number. Ans. 3. 

Reduction of Complex Fractions 

Process. Reduce first all whole numbers, if any, and 
all mixed numbers, if any, to improper fractions, and 



84 A HAND BOOK FOR MECHANICS 

then reduce all compound fractions, if any, to single 
fractions. Then divide the numerator of the complex 
fraction by the denominator, according to the rule for 
the division of fractions. 

3. 

Q 

Example. Reduce 3 toa single fraction. 

4 

This is an example of a complex fraction having for 
both terms single fractions. Knowing that the numera- 
tor of a fraction is the dividend and the denominator is 
the divisor, all that will be required to solve the above 
problem is to divide the numerator f by the denominator 

3 

4' 



a Thus 



33 34121 

8 '4~83~24~2 Am ' 



3 
Example. Reduce — to a single fraction. Thus: » 

3_ jf = 3_453jj^24 

5| V 1 " 8 1 45 45 

The above example shows that the whole numbers and 
mixed numbers were first reduced to improper fractions. 
We then proceed with the operation as in former example. 

3 

Example. Reduce L \ 3 to a single fraction. Thus: 

f f 3 . 3 2 3 * 8 24 

= n X - = — = 1-fj = 1| Arts. 



ioff | 7 8 7 3 21 

As shown, the denominator \ of }, which is a compound 
fraction, is first reduced to a single fraction and then we 
proceed as before. 

Examples for Exercise 

i 
Reduce to single fraction f Ans. §. 



REDUCTION OF FRACTIONS 85 



84 
Reduce to single fraction — Ans. 11 J 

4 

33. 3 

Reduce to single fraction ^f of \ Ans. §f § 

^i, ^ 

Following now are the rules, with examples, which 
govern operations in Addition, Subtraction, Multiplica- 
tion, and Division of Vulgar Fractions. And note, when 
performing such operations, that all fractions contained 
in a problem are first reduced to their simplest form. 



Addition op Fractions 

Addition of vulgar fractions is the process of adding 
together two or more fractions, so that their value can 
be expressed in one sum. 

When adding together a number of fractions having 
the same denominator, add together the several numer- 
ators for the new numerator and reduce the fraction 
obtained in this way to its simplest form. 

ARITHMETICAL SIGNS 

As the following signs are used in performing operations 
in fractions, it is necessary that the pupil should under- 
stand them. 

The sign + is read " plus," and means that the number 
after it is to be added to the number before it; thus, 
5+ 4 are 9. 

The sign — is read "minus," and means that the number 
after it is to be subtracted from the number before it; 
thus, 9 — 4 is 5. 

The sign X is read " multiply by," and means that the 
number after it is multiplied by the number before it; 
thus, 7 X 3 are 21. 

The sign -7- is read " divide by," and means that 



86 A HAND BOOK FOR MECHANICS 

the number before it is to be divided by the number 
after it; thus, 12 -=- 6 are 2. 

The sign = is read " equal to," and means that the 
quantity after it is of the same value as the quantity 
before it; thus, 4 + 5 = 9. 



Addition of Fractions 

Example. Add together, or find the sum of, J, f , 

and f . 

Here we have three fractions to add together, all 
having the same denominator, 8. As before stated, the 
denominator tells us how many equal parts the whole 
thing has been divided into, and the numerator shows 
us how many of those parts we have. Therefore, in the 
example, we have in the first fraction 1 part of the whole 
thing, and in the second fraction we have 3 parts of 
the whole thing, and in the third fraction we have five 
parts of the whole thing. Therefore, it is evident, to find 
the number of parts contained in the three fractions, we 
add together the numerators 1, 3, and 5, the sum of 
which equals 9. The 9 thus obtained is the new nume- 
rator, and we have the fraction |, which, reduced to its 
simplest terms, equals 1J. Then, as shown, the sum of 
I, f, and f equals I or 1J. 

Example. Add together, or find the sum of, 

if, if, if, ie, ana i^ — 16 or iif J±ns. 

Here you have again a number of fractions to add 
together, all having the same denominator, and as each 
of the numerators represents a certain number of parts 
of the whole thing that is divided into 16 equal parts, 
it is evident, to get the sum of all the parts represented 
by the several numerators, these numerators are simply 



ARITHMETICAL SIGNS 87 

added together and the sum is written over the common 
denominator, and the fraction, fi, thus formed, is re- 
duced to its simplest terms and 1M is the required sum. 

Examples for Exercise 

Add together fr, tt, t 5 t, tV, tt, and H- Arts. l}f . 
Add together t 4 t, tt, tt, tt, tt, and tt- Ans. 3 t 8 t. 
Add together f , f , J-, and J. Ans. 2. 
Add together 37, s 4 t, 3 6 t, st, and st- Ans. %\. 

To bring fractions having different denominators to 
fractions having one common denominator (that is, a 
denominator common to all the fractions) : 

First, put all the denominators down in a row. 

Second, cancel all the denominators that are alike, 
except one. That is, if there are three denominators 
alike, two of them would be canceled and one would be 
left. 

For instance, if in a row of denominators there hap- 
pened to be three fives, as follows, 5, 5, 5, two of them 
would be canceled. Thus: 

and one would be left. If there were four fives, three of 
them would be canceled. 

Third, then cancel all the denominators that will 
divide another without a remainder. By that is meant, 
the smaller denominators that will divide the larger are 
canceled, the larger one which can be divided is not 
altered, the smaller ones being left out (canceled), because 
they will divide the larger. 

For instance, if a row of denominators consisted of 
the numbers, 2, 3, 7, 5, 10, 11, the 2 and 5, which will 
divide the 10, would be canceled, as shown: 

% 3, 7, f>, 10, 11 



88 A HAND BOOK FOR MECHANICS 

Having canceled all the numbers that are alike, except 
one, and struck off the numbers that will divide others 
without a remainder, the fourth thing to do is — 

Take all the remaining numbers and place them in a 
row. 

Fifth, then see if any number will divide two or more 
of them (not less than two); if so, divide by it and place 
in the quotient those numbers which cannot be divided. 

For instance, if the numbers left were 12, 7, 16, 5, 
they would be placed in a row as follows: 

12, 7, 16, 5 

It is seen that the number 4 will divide two of these 
numbers, 12 and 16, without any remainder, so divide 
by it and also place in the quotient the numbers 7 and 5, 
which cannot be divided by 4. Thus: 

4 )12, 7, 16, 5 
3, 7, 4, 5 

And as no two of these quotient numbers, 3, 7, 4, 5, can 
be divided by any number, we divide no further. 

Multiply all the numbers in the quotient together and 
multiply their product by the divisor. 

In this case the numbers 3, 7, 4, 5 would be multiplied 
together, and the product 420 is multiplied by the 
divisor 4. 

The product 1680 obtained is the common denominator. 
That is, 1680 is the common denominator of a number 
of fractions whose denominators are 12, 7, 16, 5. 

It will often happen that two or more of the quotient 
numbers may, in turn, be divided by some number. 
When the case is such, proceed to divide as before. 

For instance, if the remaining numbers were 12, 16, 
10, 35, we would first divide the 12 and 16 by 4 and 
place in the quotient the numbers 10 and 35, which 



ARITHMETICAL SIGNS 89 

could not be divided by 4 without any remainder, as 

follows: 

4)12, 16, 10, 35 

5 ) 3, 4, 10, 35 

3, 4, 2, 7 

Now in the quotient 3, 4, 10, 35, there are two num- 
bers, 10 and 35, which can be divided by 5, so we place 
the 5 in the divisor and proceed to divide by it, placing 
in the quotient the figures 3 and 4, which cannot be 
divided, as shown. Then, as before, multiply all the 
quotient figures together and multiply their product by 
the divisors, and the result of the multiplication is the 
common denominator. 

In this case the numbers 3, 4, 2, 7 would be multiplied 
together, and the product 168 is multiplied by the 
divisors 5 and 4, and the result of the multiplication, 
3360, is the common denominator of a number of fractions 
whose denominators are 12, 16, 10, 35. 

Lastly, divide the common denominator by the de- 
nominator of the first fraction and multiply the quotient 
by its numerator, and the product obtained is the new 
numerator required. 

Then, in short, to bring fractions having different de- 
nominators to fractions having one common denomina- 
tor, place all the denominators in a row; cancel all that 
are alike, except one. Also cancel any that will divide 
into another one without remainder. 

If there is any number that will divide two or more of 
those left, then divide by it, putting down those numbers 
that cannot be divided. Repeat this till all the numbers 
are prime numbers. 

A prime number is a number which cannot be divided 
without a remainder by any number except itself and 
unity. 



90 A HAND BOOK FOR MECHANICS 

Then multiply all the prime numbers together and 
their product by the divisor; the result will be the com- 
mon denominator for all the fractions. Then divide the 
common denominator by the denominator of the first 
fraction and multiply the quotient by its numerator; its 
product is the new numerator required. Repeat this for 
each fraction. 

Example. J, f, §, f, i, f, T 7 2, A, i, and f. Bring 
these fractions to other fractions having a common 
denominator. 

All the denominators placed in a row: 

2, 6, 3, 8, 3, 8, 12, 16, 8, 4 

There are two figures 3, cancel one of them. Next 
the 2 and the 8 (two of them) and the 4 will go into the 
16, therefore they must be canceled; the 6 and 3 also, 
because they will divide the 12. Then there only remain 
two, 12 and 16; place them as below and divide them by 
4, because 4 will divide them both without a remainder. 

% % ^ & 0, & 12, 16, ft i 
4 )12, 16 
3, 4 

Then multiply the 3 by the 4 equals 12 and multiply 
the 12 by the divisor 4 equals 48, and the 48 thus ob- 
tained is the common denominator. Lastly, bring each 
fraction to one having the denominator 48. 



15 2 3 1 5 _7_ 5 1 3 

2) 6) 3) 8) 3) 8? 12) 1^) 8) 4 



Answer: 



24 40 32 18 16 30 28 15 _6_ 3fi 

48) 4?; 48> 48) 48) 48) 4¥) 48) 48) 48 

This is done, as before explained, by dividing the com- 
mon denominator by the denominators of each of the frac- 
tions, and multiplying the quotient by the numerator. 



ARITHMETICAL SIGNS 91 

The denominator 2 of the first fraction J goes into the 
common denominator, 48, 24 times. Multiply the numer- 
ator 1 by the quotient 24, and the product 24 is the 
new numerator. Do likewise with the remaining frac- 
tions, dividing the denominator of each fraction into 
the common denominator and multiply the quotient by 
its numerator, the product being the new numerator. 
The denominators of the several fractions which are 
brought to other fractions having a common denomi- 
tor must divide the "common denominator" an even 
number of times. 

To add together fractions having different denomi- 
nators : 

First bring them to fractions having one common 
denominator, then add together all the numerators found 
by dividing the several denominators with the common 
denominator and multiplying the quotient by the nu- 
merator, the number obtained by the addition being the 
new numerator, which place on the right of the several 
fractions and underneath it place the common denomi- 
nator. 

Example. Add together, |, §, f , tu, and J. 

2 )0, 3, 4, 10, % 

3, 2, 5 
_2 

10 
J5 

30 
_2 

60 common denominator. 
48 + 40 + 45 + 42+30 205 o25 



60 60 

Add together §, §, f , and \. 



3f* = 3AAns. 



92 A HAND BOOK FOR MECHANICS 

2 )2, k ft 3 

& £> 6, 8 

2 )6, 8 

3, 4 
J3 

12 

24 common denominator 

16 + 18 + 20 + 21 _ 75 _ q3 _ Ql 
24 - 24 -^24-^s 

As stated before, compound fractions must be reduced 
to single ones, whole numbers and mixed numbers to 
improper fractions, and then all are reduced to their 
lowest terms before finding the common denominator. 

Example. Add together 4f, 5f , and 7. 

Showing the mixed and whole numbers, $ 9 , V, t, 
reduced to improper fractions, 

4 7 1 

_7 

7 
_4 

28 common denominator 

133 + 164 + 196 493 _., 
__ — — __ 1 7JL1. 

28 28 28 

As shown, the mixed numbers 4 J and 5f, and the 
whole number 7, are reduced first to the improper 
fractions, V-, \ L , t- 

The common denominator, 28, is then found, and the 
different denominators are divided into it and the quo- 
tients obtained are multiplied by the different numerators. 
By dividing the common denominator by the first de- 
nominator, 4, and multiplying the quotient 7 by the 
numerator 19, we get the product 133. By dividing the 



ARITHMETICAL SIGNS 93 

common denominator, 28, by the next denominator, 7, 
and multiplying the quotient 4 obtained by the nu- 
merator 41, we get the product 164. Then divide the 
common denominator by the last denominator, 1, and 
as 1 goes into 28 just 28 times, the quotient 28 is multi- 
plied by the numerator 7, the product of which is 196. 

As shown, the three products, 133, 164, and 196, are 
added together and their sum, which equals 493, is the 
new numerator, and it is placed on the right of the 
several fractions, with the common denominator, 28, 
underneath it, thus, 4 2 9 s 3 , which fraction is then reduced 
to its simplest form. 

Example. Add J of f to 6|. 

- of - to 6| 

When reduced to J^ 51 When reduced to im- 

single fraction 16 8 proper fraction 



Common denominator, 16 \ 
3 + 102 105 



= 6 T ^ 



16 16 

As shown, the compound fraction | of f is reduced 
to the single fraction ft, and the mixed number 6| to 
an improper fraction % 1 -. 

As shown, we then have ft and V to add together. 

Examples for Exercise 

Add together J, J, \, and f . Arts. 2J. 

Add together f, f , f , and f . Arts. 2f . 
Add together f, ft, ft, and f . Ans. 3<j|y. 
Add together \ + J + f + f + \ + It + ft. Ans. 4J. 
Add together f + I + \ + f + t 3 o + f . Ans. 5ififo. 

1 4§ 
Add together f + -~ Ans. 1 ¥ | ¥ . 

¥ lZ ¥ 



94 A HAND BOOK FOR MECHANICS 

Add § of f to | of J. Arts. 3|. 
Add 18| to 17 T V Ans. 36J. 
Add together, 4j, 5f , and 7. Ans. 17iK 

Subtraction of Vulgar Fractions 

Subtraction of vulgar fractions is the process of rinding 
the difference between fractions of different values. 

Process. Reduce all compound fractions to single 
fractions, and reduce all whole numbers and mixed num- 
bers to improper fractions, and when the denominators 
of the fractions to be subtracted one from the other are 
different, bring the fractions to others having a common 
denominator, as in addition, and subtract the smaller 
from the larger numerator. Bear in mind that the sign 
— reads minus, and means that the number after it is 
to be subtracted from the number before it. 

Example. From T V take #V. 

First, find the common denominator as in addition. 
Thus: 



Example. 





3)15, 21 








5, 7 








5 

35 








3 








105: 


= common 


denominator, 


49 - 25 24 

105 "~ 105 


Answer. 




From 


xi take ■£?. 

3)18, 21 

6, 7 

6 

42 

3 


Thus: 





126 = common denominator. 



ARITHMETICAL SIGNS 95 

Divide now, new common denominator, 126, by the 
denominator 18 of the first fraction, and multiplying the 
quotient 7, obtained by the divisor, by its numerator 11, 
you get a product of 77, which is the new numerator for 
that fraction, and by dividing the common denominator, 
126, by the denominator 21 of the next fraction, and 
multiplying the quotient by its numerator 3, you get 
the product 18, which is the new numerator for that 
fraction. You then have the numerators 77 and 18; the 
smaller being subtracted from the larger leaves 59, which 
is the new numerator, as shown. 

77 - 18 __ 59 
126 ~~ 126 

Then the difference between H and #V equals AV 
Example. From f of tt take tV of f. 

Reduce the compound fractions to single fractions. 

Proceed then, as before, by bringing the fractions thus 
found to others having a common denominator, as 
follows : 

From f of tt take tV of f • 

Reduce above to single fractions. From if take tt^- 

55 112 
55 



560 
560 

6160, common denominator. 

New numerators of the above fractions, found by 
dividing the common denominator by the numerators 
and multiplying the quotient by the numerators: 

2016- 110 J9W._ 953 
6160 0100 3080 



96 A HAND BOOK FOR MECHANICS 

Example. From 6| take J of T 9 2- 

Above reduced to its simplest terms: 

27 8 9 _ 9 

T take i of 12~16 



4 



? - £ 



£ 16, common denominator. 
New denominators. 

108 - 9 99 

16 ~16~ b ^- Ans - 

Example. From 12 take 6J. 
The above reduced to improper fractions: 

-V- take */-. 
48 - 27 21 

4 ~ 4 ~ ° 4 ~ 

As shown, the whole number 12, and the mixed number, 
6f, are reduced to improper fractions. The resulting 
fractions are then brought to others having a common 
denominator, and the smaller numerator of the fractions 
thus obtained is subtracted from the larger, and the 
difference, 21, is the new numerator, as shown. 

Example. Which is the greater, J of to, or | of t%? 

3 4 

19 3 _ 5 ft _4 

p 10 10 6 J 3 9 

3 3 

Compound fractions reduced to simplest forms: 

t 3 o and f. 
27 - 40 _ 13 
90 ~90 
Then, f of j% is the greater by If. 



ARITHMETICAL SIGNS 97 

Examples for Exercise 

From | take \ = \. Ans. 

From T \ take \ = fW Ans. 

From T °5 take T 2 T = aW Ans. 

What is the difference between f of f and 1J of |? 
ilns. f. 

Which is the greater, 3| of 2f or 8 J of If? Ans. The 
second is twice as great as the first quantity. 

What is the difference between f of t 6 3, and | of tt? 

A y, e J_01 
J1716. 1430- 

From 13 take 5|. Ans. 1\. 

What is the difference between f of 14j, and f of 9J? 
Ans. T 6 e 3 o. 

From J of 14 J take f of 9 J. Ans. 6}j?§ 

Multiplication of Vulgar Fractions 

Multiplication of fractions is the process of multiplying 
fractions together. 

Process. First bring all compound fractions to single 
fractions and reduce all mixed numbers and whole 
numbers to improper fractions. In other words, bring 
each fraction to its simplest form. Then multiply the 
numerators together for the new numerator, and the 
denominators together for the new denominator, and 
reduce the resulting fraction to its simplest form. 

As before stated, the sign X reads " multiply," and 
means that the number before it is to be multiplied by 
the number after it. 

Example. Multiply f by lfV That is: 

4 V 21 — _84_ — 21 — 3 
7 ^ 16 — 112 28 4 

As shown, the mixed number lf 5 e was brought to the 
improper fraction f£. The numerators 4 and 21 are 
then multiplied together for the new numerator, 84, and 



98 A HAND BOOK FOR MECHANICS 

the denominators are multiplied together for the new 
denominator, and the resulting fraction, tt 4 2, is reduced 
to its simplest terms, f , which is the answer. 

This same example can be worked out by canceling as 
follows: 

Multiply f by 1 T 5 F . 

1 3 

y n 4 

4 

The 4 cancels into the 16 four times, and the 7 into 
the 21 three times Then, 1x3 = 3, and 1X4 = 4. 
Ans. f . 

Example. Multiply f by to, as follows: 

5 V -3- — 15 — 1 ArtQ 
9^10 — 90 — 6 ■tltt'O. 

or by canceling: 

1 1 

^-X^ = l -Ans 

$ x ;p 6 s ' 

3 2 

It is seen by referring to the above example that 
multiplying one fraction by another is the same thing as 
reducing compound fractions to simple ones. 

Example. Multiply, 2 T V of 3f by 61 of 2 8 t. 

The above mixed numbers, when reduced to improper 
fractions : 

f* Of ¥ X - 4 # Of *. 

3 5 7 

2 1 3 

The above compound fractions, after canceling and 
being reduced to simple fractions: 



ARITHMETICAL SIGNS 



99 



15V 7 105 171 



Ans. 



or ^Xg = y= 17J. Ans. 



Then: 



2 T V of 3* X 6^ of ft = 17J 

Examples. Multiply 8| by 4. 

67 xy j* __ 268 
1 8 

Example. Multiply 82 by 

¥ X A = -W- = 67 T V Ans. 



X f = -77- = 33£. Ans. 



_9_ 
1 1 



Examples for Exercise 



Multiply 5f by 2f . Ans. 15g 3 2. 
|of|of|x|of|off. 



Arts. - 



28' 



M X 48. 



Ans. 43h 



I of 2i X I of 3i 



Ans. 2JLf. 



I of I of B X I of 4|. Ans. 1J. 
4y X lio- Ans. 235. 



3 8 

12 X 5' 



Ans 



^5 



Division of Vulgar Fractions 

Division of fractions is the process of obtaining the 
number of times one fraction is contained in another. 

A fraction may be divided by either dividing its 
numerator or by multiplying its denominator. 

Example. Divide t 4 t by 2. 

Here we have t 4 t of a whole thing which is composed 

L OF C. 



100 A HAND BOOK FOR MECHANICS 

of 11 equal parts, and those four one-elevenths (i*r)are 
to be divided by 2. It is here evident that by dividing 
the numerator 4 of the fraction t 4 t by 2, and placing the 
quotient over the denominator, the resulting fraction 
will be the required answer. Thus: t 4 t divided by 2 = tV 
Ans. 

Instead, now, of dividing the numerator 4 of the frac- 
tion tt, by 2, multiply the denominator 11 by 2 and 
place the product underneath the numerator and reduce 
the resulting fraction to its lowest terms. The same result 
will be obtained as was obtained by dividing the nu- 
merator 4 by 2. Thus: 

4 4 2 

IT x2 = ?2 = ir Ans - 

It is here fully demonstrated that either dividing the 
numerator or multiplying the denominator divides a 
fraction; however, for convenience, operations in division 
are performed by inverting the terms of the division, 
that is, by placing the denominator above the horizontal 
line and the numerator below it, as shown in following 
examples. 

It will be remembered that the sign -=- reads " divide 
by," and means that the number before it is to be divided 
by the number after it. Thus: 



_4_ 

1 1 



~ _ — y y X. ■" — 22 — 11* AnSm 



Example. Divide tt by 5. Thus: 

t 8 t - 5 = ty X 5 = ft. Ans. 

Example. Divide f by 6. Thus: 

2^6=2x6 = _ 2 _ = ¥ i y> A nSt 

Example. Divide 14 by f. Thus: 



ARITHMETICAL SIGNS 101 

14 + f = 14 X i = 98 

3)98(32|. Arts. 
9^ 

8 

A 

2 

3 

Example. Divide 32 by t 9 t- Thus: 

~9)352(39£. Arts. 

27 

82 

81 

1_ 

9 

All mixed numbers and whole numbers must be re- 
duced to improper fractions, and all compound fractions 
reduced to single fractions, before dividing. 

Example. Divide If by 1J. 

Reduce first the mixed numbers to improper fractions, 
then proceed to divide. Thus: 

13 _^ 11 

Above reduced to improper fractions. 

1 -J- £L 
4 • 8 

Then 



7^? = 7 x 8 = 56 

4 ' 8 4 9 36 9 

Example. Divide f by 1|, that is: 

3 . 9 — 3 V 8 24--2 Avtd 

Example. Divide f of 2 by f of 7. 

Reduce first all whole numbers to improper fractions. 



102 A HAND BOOK FOR MECHANICS 

Thus: 

| of f + | of 1 

Reduce, then, above compound fractions to single 
fractions. Thus: 

| of f - | of 1 

Above reduced to single fractions: 



TViAn 


6 _i_ 2 1 
4 * 4 




_L IlfcJIl 


6 -21 6 V 4 — 24 -_ 2 

4 ' 4 4 A 21 84 7- 


Ans. 


Example. 


Divide 4f of \% by 3f of 3^. 


That is: 


That is: 


30 n fl4 _t_ 1 5 n f 1 6 
"7 U1 15 • 4 U1 5 




WlJa 

i 


3 4 
_M of W_4 x ±_±_ 
t 1 12 " 12 ~ 


-. Ans. 

o 



Canceling, as shown above, reduces the dividend to \ 
and the divisor to V. 

Examples for Exercise 

\ + | Ans. f . 
| -T- J Ans. 2. 
| -f- i Ans. 1J. 
3J -s- f Ans. 6J. 

of 12i - i of 84. Ans. 1031. 



3 
TT '5 



£ of 81. 

« 2 

5# ~^" ^5- Ans. ^6ff« 
4f of if -s- 3| of 3J. Ans. f£. 
§ of | -^ | of |. Ans. 18|. 
| of f of f ^ f of f of T V Ans. 3ff. 



DECIMAL FRACTIONS 103 

DECIMAL FRACTIONS 

The term "decimal" is derived from the Latin word 
deci, or decimus, which means ten or tenth. 

The decimal system of figures is very old and in all 
probability originated from the use of the ten fingers as 
helps to count. 

It is supposed that in India there was in use, as early 
as a.d. 525, an imperfect form of our present decimal 
system. The system was not used in Europe till about 
1200, when it was introduced through" the Arabians. 
According to the decimal system, ten or some multiple 
of ten units of a lower denomination make one unit of 
the next higher denomination. 

Our present money system, for instance, is a decimal 
system, because: 

10 mills make one cent. 
10 cents make one dime. 
10 dimes make one dollar. 
10 dollars make one eagle. 

According to the decimal system, a dot, thus (.), 
called the decimal point, separates the unit from the 
fractional parts of an expression. Thus, in the following 
decimal expression: 

354.5 

the 354 represents units and the .5 represents the frac- 
tional parts of a unit. A decimal fraction is one whose 
denominator is always 10 or 100 or 1000, or some other 
power (as it is called) of 10, but its numerator may be 
any number. 

For example, T V, T £o, Woo, to, too, are all decimal 
fractions. 

Strictly speaking, there should be as many figures in 



104 A HAND BOOK FOR MECHANICS 

the numerator as there are cyphers in the denominator; 
and if there are not as many, they may be made so by 
attaching cyphers to the left of the given numerator. 
For example, the five fractions above would be: 

JL JUL 001 3 J)_6_ 

10) 100) 1000) 10) 100* 

This being the case, the denominator can be done away 
with, a dot, thus (.), being placed before the numerator. 

A decimal fraction, then, is always represented by 
one number having a dot before it ; that is : 

"to is written .1 and is in value equal to one tenth of 
a whole number. 

to is written .7, and is in value equal to seven tenths 
of a whole number. 

tVo is written .01, and is in value equal to one hundredth 
of a whole number. 

toVo is written .001, and is in value equal to one 
thousandth of a whole number. 

So it will be seen that, in decimals, by placing a figure 
one place to the right makes it a tenth of what it was 
before, just as in whole numbers. Thus: 

1. is one whole thing. 

.1 is one tenth whole thing. 

.01 is one hundredth whole thing. 

.001 is one thousandth whole thing, etc. 

If the fractions have a numerator other than 1, thus: 
T % is written .5, T Vo is written .27, and f\ 7 o 2 o is written 
.372. 

The first place on the right of the decimal point is 
always tenths, the second place on the right of the 
decimal point is always hundredths, and the third place 
on the right of the decimal point is always thousandths. 

Knowing, then, that a decimal fraction is expressed 
without its denominator, by means of the decimal point, 



DECIMAL FRACTIONS 105 

and that it may be expressed in this way because the 
denominator of a decimal fraction is always 1, with as 
many places annexed as there are places in the deci- 
mal, proceed to express the following in decimal form: 

(1) T Vo Ans. .85 ; (2) T U Arts. .06 ; (3) T c°o o Ans. .009 ; 
(4) T Vo Ans. .75 ; (5) ftVo Ans. .201 ; (6) 349 WoW 
Ans. 349.000019 ; (7) 12 tVo 2 o Ans. 12.342 ; (8) T o oVWo 
Ans. .000012. 



To Bring a Decimal Fraction to a Vulgar 
Fraction 

Process. Put down the given decimal as a numerator, 
leave out the decimal point, then place underneath it a 
denominator 1 with as many cyphers after it as there 
are places in the given decimal. 

For instance, .25 to a vulgar fraction. Thus: 

100 4* -tli I*' 

The operation was performed by omitting the decimal 
point and placing the denominator underneath the given 
numerator, then changing the fraction to its lowest terms. 

Example. Bring .875 to a vulgar fraction. Thus: 

_8_7_5_ — 17 5 3 5 — 7_ A ™ Q 

1000 200 40 8 ^-'^. 

Example. Bring .87500 to a vulgar fraction. Thus: 

87^00 — J7 500 — lioOO 3 000 __ 7.0 — 7 A ~, Q 

•°« <J\j\j iooooo 20000 — 4000 — so 8 ^ Lf i'0. 

Which is the same as the preceding example and 
demonstrates that cyphers on the right of a decimal 
have no value. You may add as many cyphers to the 
right as you please, or if any are given (note the fact) 
you may leave them off without affecting the value of 
the decimal. Thus: 



120 = .1200 and .12 = .120 or .1200 



106 A HAND BOOK FOR MECHANICS 

Examples for Exercise 
Bring the following decimal fractions to vulgar fractions. 



1. 


.025. 


Ans. 


i 

40- 


5. 


120.6. 


Ans. 


120f. 


2. 


.476. 


Ans. 


129 

250- 


6. 


.581. 


Ans. 


58 1 

T000- 


3. 


.50. 


Ans. 


1 
2' 


7. 


.01395. 


Ans. 


2 79 
20007T. 


4. 


.050. 


Ans. 


1 

20- 


8. 


300.98. 


Ans. 


300M. 



To Bring a Vulgar Fraction to a Decimal 
Fraction 

Annex any number of decimal cyphers to the numerator 
and divide by the denominator, and point off as many 
decimal figures in the quotient as there are cyphers 
annexed to the numerator. 

Example. Bring J to a decimal fraction. 

Proceed by placing a cypher on the right of the nu- 
merator 1, then divide this by the denominator 4, as 
follows, saying 4 into 10 goes 2 and 2 over. Annex then 
another cypher and divide by 4, again saying, 4 into 20 
goes 5 times. Then mark off as many decimal figures in 
the quotient as there were cyphers annexed to the nu- 
merator. Thus: 

4 )1-00 

.25 Ans. 

Example. Bring II to a decimal fraction. Thus: 

16)15.00(.9375 Ans. 
144 
60 
48 
120 
112 
80 
80 



DECIMAL FRACTIONS 107 

Sometimes, in reducing a vulgar fraction to a decimal 
fraction, the quotient never comes to an end, but the 
same figure keeps repeating itself, as J = .16666, etc., 
without end. This is called a repeating decimal and is 
written .16. The dot over the 6 represents that the 6 is 
a repeater. 

Sometimes, instead of one figure repeating itself, 
as above, you will find two or more doing so, as \ = 
.142857142857142857, etc. Thus: 

+)1.00000000000(.142857142857 
7_ 
30 
28 
20 
14 
60 
56 
40 
35 
50 
49 
10 
7_ 
30 
28 
20 
14 
60 
56 
40 
35 
50 
49 
10 



108 



A IT AND BOOK FOR MECHANICS 



Here it will be seen that the figures 142857 keep 
coming over and over, and this would continue without 
end. This is called a circulating decimal and is written 
thus, .142857, having a dot placed over the first and 
last figure of the set of figures which keep repeating. 

Then, \ = .142857142857142857 etc., and is expressed, 
.142857. 

It is not often necessary to carry decimals more than 
six places, thus, .534678. 

Reduce to decimal fractions the following: (1) § Ans. .75; 
(2) i Ans. .5; (3) g Ans. .375; (4) | Ans. .875; (5) A 
Ans. .125; (6) H Ans. .6875; (7) M Ans. .7; (8) ft 
Ans. .32; (9) f Ans. .83; (10) H Ans. .916. 



To Reduce a Mixed Number to a Decimal 
Fraction 

First, reduce the mixed number to an improper frac- 
tion, then reduce the improper fraction to a decimal 
fraction. 



Example. Reduce 4^? to a decimal. 

4^)257.000000(4.015625 
256 
100 

360 
320 



Ans. 



400 
384 



Then, 4& = V 



2 5J 
64 



160 

128 
320 
320 
4.015625. 



Example. 



DECIMAL FRACTIONS 

Reduce 11£ to a decimal. 

11J = 23 

2)23.0(11.5 
22 



109 



10 
10 



Then 11J= V = 11.5 Ans. 

To reduce a decimal fraction to any particular vulgar 
fraction, multiply the given decimal by the denominator 
you wish to bring it to, then mark off as many decimals, 
from right to left, as there are places in the given decimal, 
and whatever number is to the left of the decimal point 
is the required numerator. 

Example. How many sixteenths are there in .198? 

.198 
16 





1188 
198 




3.168 Ans. (a little over f\) 


Example. 


How many halves in .5? 




.5 
2 

1.0 Ans. 1 


Example. 


How many fourths in .75? 




.75 
4 




3.00 Ans. 3 (three fourths). 



Addition of Decimals 

Process. Place the quantities in such manner that 
the decimal point of one line shall be directly under that 



110 A HAND BOOK FOR MECHANICS 

of every other line, and arrange units under units, tenths 
under tenths, hundredths under hundredths. Then add 
as in whole numbers, placing the decimal point in the 
sum directly underneath those above. 

Example. Add together, 18; 74; 3.98046; 273.4; 
15.0071; .04527. Thus: 

18 
74 

3.9S046 
273.4 
15.0071 
.04527 
384.43283 

Examples for Exercise 

Add together: 

(1) 30.034; 2135.2; 27.03; 2.34071. Ans. 2194.60471. 

(2) 15.17; 2.3876; 403.20104; 12.5. Ans. 433.25864. 

(3) 246.03; 00.674; 5.012; 15.999. Ans. 267.715. 

(4) Find the sum of 327.05 + 12.491 + 16.25 + 642.75 
Ans. 998.541. 

(5) What is the sum of 5.0327 + 491.12 + 25.16 + 
75.2. Ans. 596.5127. 

Subtraction of Decimal Fractions 

Process. Place the less number under the greater, 
with decimal point under decimal point, as in addition. 
If one line has more decimal figures than the other, put 
noughts at the end of the one that is deficient till they 
are equal, then subtract, as in single subtraction, and 
point off as many places for decimals as there are in the 
number which contains the greater number of decimals. 



DECIMAL FRACTIONS 11 I 

Example. From 232.5003 take 76.8973. ' 

232.5003 
76.8973 
155.6030 Ans. 

As shown, the less number is written under the greater, 
arranging units under units, tenths under tenths, hun- 
dredths under hundredths, etc. Then the less number 
is subtracted from the greater, as in whole numbers. 

Example. From 4.17 take 2.893652. 

4.170000 
2.893652 

1.276348 Ans. 

Here, as shown, one line of decimal figures is greater than 
the other, and at the end of the line that was deficient, 
noughts were placed till the number of decimal places were 
made equal in each line. Then we subtract as before. 

Example. From 100 take .001. 

100.000 
.001 

99.999 Ans. 

Examples for Exercise 

1. From 24.36 take 16.054267. Ans. 8.305733, 

2. From .42052 take .1360076. Ans. .2845124. 

3. Find the value of 467 - 43.6054 Ans. 423.3946. 

4. Find the value of 12.036 - 5.61. Ans. 6.426 

5. From 43.123 take 8.58195. Ans. 34.54105. 

Multiplication of Decimals 

Process. Multiply as in common multiplication, with- 
out taking notice of the decimal points. Add up and so 
get the product. 



112 A HAND BOOK FOR MECHANICS 

Point off as many figures for decimals in the product 
as there are decimals in the multiplicand and multiplier. 
There must be as many decimal places in the product as the 
sum of the decimals in the multiplicand and. multiplier. 

Example. Multiply 45.32 by 6.531. 

45.32 
6.531 
4532 
13596 
22660 
27192 



295.98492 



The product first stood 29598492, but as there are 
altogether five decimal figures in the question (two in 
the multiplicand and three in the multiplier), we count 
five, beginning at the last figure 2, and place a decimal 
point before the figure that stands at the fifth place. 
The answer is 295.98492. 

It is sometimes found that there are not as many 
figures in the product as there are decimal figures in the 
multiplicand and multiplier. In such cases, put a suffi- 
cient number of cyphers on the left of the product to 
make the number of decimal places equal to the sum of 
the decimal places in the multiplicand and multiplier. 

Example. Multiply .00052 by .0402. 

.00052 

. 0402 

104 

208 
20904 

Here it is seen that the product, 20904, contains five 
figures only, and that there are nine decimal places in the 



DECIMAL FRACTIONS 113 

example. We therefore place four cyphers on the left of 
the product, making it .000020904, which is the answer. 

Example. Multiply 4.12 by .013. 

4.12 
.013 



1236 
412 



.05356 Arts. 

Examples for Exercise 

1. Multiply 345.2 by 32.14. Ans. 11094.728. 

2. Multiply 42.545 by 3.42. Ans. 145.50390. 

3. Multiply 2084 by .3055. Ans. 636.6620. 

4. Multiply .5652 by .0025. Ans. .00141300. 

5. Multiply .0004 by .00304. Ans. .000001216. 

6. Multiply .0002 by .00101. Ans. .000000202. 

Division of Decimals 

Process. When the divisor is a whole number, divide 
as in simple division, and when you come to the decimal 
point in the dividend, place a point in the quotient. 

Example. Divide 382.7838 by 2. 

2 )382.7838 
191.3919 

Example. Divide 1537.27 by 8. 

8)1537.27 
192.15875 

As shown, 8 into 15 goes 1 time and 7 over, then 8 
into 73 goes 9 times and 1 over, then 8 into 17 goes 2 
times and 1 over, 8 into 12 goes 1 time and 4 over; 
then after saying 8 into 47 goes 5 times and 7 over 



114 A HAND BOOK FOR MECHANICS 

make this 7 into 70; 8 into 70 goes 8 and 6 over; 8 into 
60 goes 7 times and 4 over, 8 into 40 goes 5 times. 

Example. Divide 144.144 by 12. 

12)144.144 

12.012 Arts. 

Here we say, 12 into 144 goes 12 times, which place 
in the quotient. Then on the right of it, and under- 
neath the decimal point in the dividend, place the decimal 
point. We then say, 12 into 1 goes no times, which 
place in the quotient, then 12 into 144 goes 12 times. 

Example. Divide 38.4256 by 12. 

12)38.4256(3.20213 
36 
24 
24 



25 
24 

16 
12 

40 

36 

4 

Here, after the 6, the last decimal figure in the dividend, 
is brought down, it goes 2 times and the remainder is 4, 
to this attach a cypher and let it go on again as far as it is 
thought necessary. Note here that the last decimal figure, 
3, of the quotient, will keep repeating over and over, and, 
as before stated, a decimal figure that repeats is called a 
repeater and is expressed by placing a dot over it. There- 
fore, over the 3 in the quotient is placed a dot, thus, 3. 

When the number of decimal figures in the divisor is 
less than the number of decimal figures in the dividend, 



DECIMAL FRACTIONS 115 

divide without taking notice of the decimals; then sub- 
tract the number of decimals in the divisor from the 
decimal figures in the dividend; the remainder will be 
the number of decimals to mark off in the quotient. 
That is, if, for instance, there was one decimal figure in 
the divisor and three decimal figures in the dividend, 
after dividing we would subtract the one decimal figure 
of the divisor from the three decimal figures of the divi- 
dend, and the remainder two would be the number of 
decimal places to mark off in the quotient. 

Example. Divide 153.4036 by .4. 

.4 )153.4036 
383.509 

Here there is one decimal place in the divisor and 
four decimal places in the dividend. So we say, 1 from 
4 leaves 3, then mark off 3 decimals in the answer. 

When marking off the decimal places in the quotient, 
mark off from right to left a sufficient number of figures 
so that those figures with the decimal places in the 
given divisor will together equal the number of decimal 
places in the given dividend. That is, the number of 
decimal places in the quotient and divisor together must 
be equal in number to the number of decimal places in 
the dividend. 

Example. Divide 182.6025 by .25. 

.25)182.6025(730.41 
175 
76 
75 
102 
100 
25 
25 



116 A HAND BOOK FOR MECHANICS 

Here we have four decimal places in the dividend and 
two decimal places in the divisor; therefore, we mark off 
in the quotient, from right to left, two decimal places. 
We then have in the quotient and divisor together the 
same number of decimal places as there are decimal 
places in the dividend. 

Example. Divide 136.875 by 37.5. 

37.5)136.875(3.65 
1125 
2437 
2250 



1875 
1875 



Example. Divide 136.875 by 3.75. 

3.75)136.875(36.5 
1125 
2437 
2250 



1875 
1875 



It will sometimes happen that the number of decimals 
in the quotient and divisor together will not equal the 
number of decimals in the dividend. In such a case, 
prefix a sufficient number of cyphers to the figures in 
the quotient to supply the deficit, as shown in the fol- 
lowing example. 

Divide .525 by 7.5. 

7.5).525(.07 
525 

In the above example we place a cypher before the 
quotient figure 7, which removes it to the place of hun- 



DECIMAL FRACTIONS 117 

dredths and then makes the decimal places of the divisor 
and quotient together equal to the three decimal places 
of the dividend. 

Example. Divide .0123825 by .75. 

.75).0123825(.01651 
75 
488 
450 



382 

375 

75 



Here we divided as in whole numbers and have in the 
quotient four figures, 1651, to which we prefixed a cypher 
so that the decimal places of the quotient together with 
the decimal places of the divisor will equal the number 
of decimal places of the dividend. When there are more 
decimal figures in the divisor than there are in the divi- 
dend, annex as many cyphers to the decimals of the 
dividend as will make them equal to the number of 
decimals of the divisor. That is, equalize the decimals 
by attaching cyphers to the dividend, so that the divisor 
and dividend will contain the same number of decimals. 
Then leave out the decimal points altogether and divide 
as in simple division, and the quotient will be a whole 
number. If there is a remainder after this, attach a cypher 
to it and again divide; this will give the first decimal 
figure of the quotient ; to the remainder again attach a 
cypher; again divide for the second decimal figure, and so 
on, as far as it is thought necessary. 

Example. Divide 2.2 by .550. 

First equalize the decimal figures, thus, 2.200 by .550. 
Then leave out the decimal points, and divide thus: 



118 A HAND BOOK FOR MECHANICS 

550)2200(4 Ans. 
2200 

Ans. 4 whole numbers. 
Example. Divide 1562.5 by .00025. 

First equalize the decimal places of the divisor and 
dividend by annexing cyphers to the dividend, thus, 
1562.50000 by .00025; leave out the decimal points and 
divide as in simple division. 

00025)156250000(6250000 Ans. 
150 
62 
50 



125 
125 



Example. Divide 147.24 by .84625. 

Equalize 147.24000 by .84625. 

84625)14724000(173.99, etc. 
84625 
626150 
592375 



337750 
253875 
838750 
761625 



771250 
761625 

Referring to the above example we see that the divisor 
goes into the dividend once, then 7 times, and then 
3 times, and as there are no more figures left to bring 
down, these 173 are whole numbers. To find the deci- 
mals, attach a cypher to the remainder, 83875, and it 



DECIMAL FRACTIONS 119 

goes 9 times. This is put in the quotient as .9 (note the 
decimal point); to the next remainder, 77125, attach 
another cypher and it goes 9 times again. Put this 9 
after the former one, attach another cypher to the re- 
mainder, if thought necessary, and continue as far as 
you please. However, as before stated, it is not usually 
necessary to carry decimals more than six places. 

Examples foe Exercise 

Divide 72.348 by 6. Ans. 12.058. 
Divide 468.6520 by 5. Ans. 93.7304. 
Divide 144.12 by 12. Ans. 12.01. 
Divide 33.9316 by 16.4. Ans. 2.069. 
Divide 176.24 by .6725. Ans. 262.0669. 
Divide 983. by 6.6. Ans. 148.93. 
Divide 1562.5 by .0025. Ans. 625000. 
Divide 2944.8 by .21312. Ans. 13817.567, etc. 
Divide 144.144 by 12.12. Ans. 11.8939, etc. 
Divide 147.24 by .84625. Ans. 173.99, etc. 
Divide 87916.05 by .88. Ans. 99904.6022, etc. 
Divide 2.28125 by 62.5. Ans. .0365. 
Divide .228125 by 62.5. Ans. .00365. 

PROPORTION, OR THE RULE OF THREE 

Proportion is the process of finding the relation of one 
thing to another, in respect to size, quantity, capacity, 
or degree. In proportion there are three quantities given, 
and with the three given quantities we are to find a 
fourth, which quantity is to bear the same relation to 
the third quantity as the first quantity bears to the 
second. Then, according to theory, proportion is the 
rule by which we are enabled to find the fourth propor- 
tional to three given numbers. That is, a fourth number 



120 A HAND BOOK FOR MECHANICS 

which will bear the same relation or ratio to the third 
number as the first number bears to the second. 

The term "ratio" is used more or less in working out 
problems by the process of proportion or the rule of 
three, and ratio means the relation between two similar 
things in respect to quantity or size; that is, the rela- 
tion between two like things, in respect to how many 
times one makes so many times the other, and is the 
same as quotient, which expresses how many times one 
quantity contains another. Then it is necessary to have 
two numbers to form a ratio, because ratio is the quotient 
arising from the division of one number by another. 
Thus, the ratio of 8 to 4 is 2. 

Of the two numbers required to form a ratio, the first 
is called the antecedent and the last the consequent. 

Then, in the above example, 8 is the antecedent and 4 
is the consequent. 

A ratio may be either direct or inverse. A direct ratio 
is when the antecedent is divided by the consequent, 
that is, when the first term is divided by the second; and 
an inverse ratio is when the consequent is divided by 
the antecedent, that is, when the second term is divided 
by the first. Then the direct ratio of 8 to 4 is 8 ^ 4 
and the inverse ratio of 8 to 4 is 4 -~- 8. 

Proportion, then, enables us to find a fourth number 
when there are three numbers given that will be the same 
ratio to the third term that the first term is to the second. 

A ratio is usually expressed by two dots, thus (:), and 
a proportion is usually expressed by four dots, thus (::). 

The sign : reads "is to." 

And the sign : : reads "as." 

Then when performing examples in proportion, write 
same thus: 

3:6::6: 12 



PROPORTION, OR THE RULE OF THREE 121 

For instance, 3 is to 6 as 8 is to what? 

Here there are three given quantities, 3, 6, and 8, and 
we have to find a fourth quantity that will be the same 
relation to 8 that 3 is to 6. The rule is to multiply the 
second term by the third term and divide the product 
by the first term, and the quotient will be the fourth 
term required. 

Example. 3 is to 6 as 8 is to what? 

3:6::8:16 
J6 

3)48 



16 required term which is 
placed above, as shown. Then 3 is to 6 as 8 is to 16. 



Example. 


6 


is to 3 as 


16 is t( 






6:3::16:8 
3 

6)48 
8 


Then 6 is to 3 


as 


16 is to 8. 





Here, as before, multiply the second term by the third, 
and divide the product by the first term and the quotient 
is the required number. 

Example. 14 : 8 : : 6 : what ? 

Thus: 14:8::6:3f Ana. 
8 
14)48(3? 
42 

14 ~ 7 



122 A HAND BOOK FOR MECHANICS 

Example. 8: 14:: 5: what ? 

Thus: 8:14::5:8| Arts. 
14 
20 
5 



8)70(8| 
64 
J5 3_ 

8 ~ 4 

Example. 10 : 20 : : 12 : what ? 

Thus: 10 :20 :: 12 : 24 
20 
10)240 

24 Ans. 
Example. 20 : 10 : : 24 : what ? 

Thus: 20: 10:: 24: 12 
10 



20)240(12 Ans. 
20 

40 
40 

It is now seen in all cases there are three quantities 
given, as stated before, and with the three given quan- 
tities we are to find a fourth. And we have seen that 
by multiplying the second term by the fourth term, and 
dividing the product by the first term, the quotient is 
the required fourth term. Now, in all cases, two of 
the three given terms will be of the same denomina- 
tions; that is, two of the terms will always represent 
the same kind of a thing, and the two terms of a like 
denomination will always be the first and second 
terms of the proportion, and the other given term 



PROPORTION, OR THE RULE OF THREE 123 

will have the same denomination as the required answer, 
and this term will always be the third term in the pro- 
portion. 

To know which is to be the first and which the sec- 
ond term of the proportion: 

If the answer is to be greater than the third term, the 
greater of the like terms must be the second term of the 
proportion, but if the answer is to be smaller than the 
third term, the smaller of the like terms must be the second 
term of the proportion. 

Example. If four men can earn $10 per week, how 
much will 8 men earn in the same length of time? 

Here there are two terms of the same denomination, 
4 men and 8 men, and one term of a different de- 
nomination, $10. The unlike denomination being the 
third term of the proportion, the $10 must be the third 
term. 

To find the second term — will 8 men earn more than 
4 men? 

Yes. Then the 8 is the second term, consequently 4 is 
the first term. The proportion, therefore, is 

4: 8:: 10: Ans. 
8 
4)80 

20 dollars. Ans. 

If a man can earn $17 in 7 days, how many days will 
it take him to earn $170? 

Here the 7 days will be the third term, and as the 
answer must be more than that, the larger number of 
dollars must be the second term and the smaller number 
of dollars the first term of the proportion. 

Then, as 17 : 170 : : 7 : the answer. 



124 A HAND BOOK FOR MECHANICS 

17: 170:: 

7 



17)1190(70 days 
119 

Arts. 70 days. 

Example. If a man can earn $2.50 in one day, how 
long will it take him to earn $170? 

Here the third term is 1 day, and as it will take him 
longer to earn $170 than $2.50 (2.50), the second term of 
the proportion must be $170, and consequently the first 
term is $2.50. 

Then: 2.50: 170.00:: 1: the answer. 

1 

250)17000(68 days. Arts. 
1500 
2000 
2000 



Example. If a ship sails 26 knots in 4 hours, how 
long will it take her to sail 650, at the same rate of sailing? 
Here the 4 hours will be the third term, and as the answer 
must be more than that, the larger number of knots 
must be the second term and the smaller number the 
first term. 

Then as 26: 650:: 4: the answer. 

4 



26)2600(100 hours. Ans. 
26 
00 
or 4 days, 4 hours, of 24 hours per day. 

Example. If 8 men can do a certain amount of work 
in 18 working days, how long should it take 12 men to 
do it? 



PROPORTION, OR THE RULE OF THREE 125 

It is plainly seen that it will take 12 men less time to 
accomplish the work than it would 8 men, therefore, the 
less number, 8, must be the second term. Then 

12: 8:: 18: the answer. 



12 )144 

12 days. Ans. 

Example. If I pay $3 for twelve yards of cloth, how 
much will 8 yards cost? 

It will manifestly take less to buy 8 yards than it takes 
to buy 10 yards, therefore, 8 must be the second term of 
the proportion. Then 

12: 8:: 3.00: the answer. 



12)2400 

2.00 dollars. Ans. 

Examples for Exercise 

1. If 12 men earn $96 per week, what would be the 
wages of 3 men for the same time? Ans. $24. 

2. If 6 paces taken by a man equals 5 yards, what is 
the length of a plank that is 240 paces? Ans. 200 yards. 

3. If my expenses are $12 per week, how long will 
$465 last me? Ans. 38f weeks, or 271J days. 

4. If 72 men cut 96 cords of wood per day, how many 
cords will 5 men cut? Ans. 6 J cords. 

5. If 10 men can do a certain amount of work in 50 
days, how long will it take for 7 men to do the same 
amount of work? Ans. 71y days. 

6. If 10 men engage to complete a certain amount of 
work in 50 days, but 3 of them are taken sick, how long 
will it take the rest to complete the work? Ans. 71y days. 



126 A HAND BOOK FOR MECHANICS 

7. If a house can be built in 14 days by a certain 
number of men, working 12 hours a day, how many 
days will it take them to build it if they work 8 hours a 
day? Ans. 21 days. 

8. If a house can be built in 14 days by a certain 
number of men working 10 hours per day, how many 
days will it take them to complete it if they work 12 
hours per day? Ans. 11§ days. 

9. If 6 men can do a certain amount of work in 14 
days by working 12 hours per day, how many days will 
it take them to do the work if they work 8 hours per 
day? Ans. 21 days. 

10. If 8 men can build a house in 90 days by working 
8 hours per day, how long will it take them to build it 
if they work 12 hours per day? Ans. 60 days. 

Note. Go over these examples thoroughly and care- 
fully. Understand the sense of the examples and it will 
be a simple matter to solve them, and in order to sim- 
plify future work care must be taken that this chapter 
is thoroughly understood. 

Compound Proportion 

To compound anything is the putting together two 
or more elements, and by the process of compound pro- 
portion we are enabled to perform questions by putting 
together two or more elements (compounding) that would 
require two or more operations to perform if performed 
by the process of simple proportion. 

In compound proportion there are always two or more 
elements to be considered, as men, dollars, and days, for 
instance. We will call each denomination, for conven- 
ience, an element. In arranging numbers in the pro- 



PROPORTION, OR THE RULE OF THREE 127 

portion, the third term of the proportion is always like 
the answer (the reason for this is more fully shown 
below)/ Only one of the elements must be considered 
at a time and the different quantities in this element 
must be arranged in the proportion as if the answer 
depended upon this element alone. For instance, if 8 
men can earn $128 in 16 days, how much can 24 men 
earn in 6 days? 

Here the different elements are dollars, men, and days. 
Now the question is, to find the number of dollars that 
can be earned by a certain number of men in a certain 
number of days, knowing a certain number of men can 
earn $128 in a given number of days. Therefore, dollars 
is the third term, because, as stated above, that number, 
or element, of the question which is like the answer is 
made the third term of the proportion. 

Now take the other elements of the question. First 
take men and arrange the proportion as if the answer 
depended upon men only, saying, if 8 men can earn 
$128, how many dollars can 24 men earn? More. There- 
fore, 24 is the second term of the proportion and 8 is the 
first term, and as we know 128 to be the third term we 
have: 

8:24:: 128 

Now take the next element in the question, which is 
days, and arrange in the proportion, the different quan- 
tities of this element as if the answer depended on it only, 
saying, if $128 can be earned in 16 days, how many can 
be earned in 6 days? Less will be earned in 6 than in 
16 days, consequently 6 is the second term of the propor- 
tion in arranging this element. Thus: 

16:6:: 

Now arrange the different elements as follows, placing 



128 A HAXD BOOK FOR MECHANICS 

the first and second term of each element under the 
first and second term of the other. Thus: 

8: 24:: $128: the answer $144 
16: 6:: 

Now find the continued product of the second and 
third terms by multiplying 128 X 24 X 6 and divide 
the product by the continued product of the first terms, 
that is, 8 X 16, and the quotient is the required answer. 

Then: 



6 X 24 X 128 


18432 


16 X 8 


128)18432(144 Ans 
128 




563 




512 


. 


512 




512 



Example. If 4 horses eat 21 bushels of grain in 12 
days, how many bushels will last 16 horses 24 days? 

4: 16:: 
12: 24:: 21: answer 168 bushels 

16 X 24 X 21 8064 



4 X 12 48 



= 168 bushels. Ans. 



As the answer is in bushels, we make the 21 bushels 
the third term. If the answer depended only on the 
number of horses, it would be larger than the third term, 
as 16 horses will eat more than 4 horses. Then 16 is the 
second term and 4 the first term, and if the answer de- 
pended only on the number of days, it would also be 
larger than the answer, because, if 21 bushels are eaten 
in 12 days, a greater number of bushels will be eaten in 
24 days. Hence, the second ratio, 12:24. Then, as 
before, we divide- the product of the two second terms 



PROPORTION, OR THE RULE OF THREE 129 

multiplied by the third terms by the product of the two 
first terms, and the quotient is the required answer. 

Example. If 8 horses mow a field of 128 acres in 16 
days, how many acres will 24 horses mow in 6 days? 

Here the different elements are horses, acres, and days, 
and the proposition is to find the number of acres that 
can be mowed by 24 horses in 6 days. If a given number 
of acres can be mowed by 8 horses in a given number of 
days, acres then is the third term of the proportion, 
because it is like the answer. Now take the other ele- 
ments in the proportion and arrange them in the pro- 
portion as if the answer depended upon the element 
taken only. 

We will first take horses and inquire, if 8 horses can 
mow 128 acres in 16 days, how many acres will 24 horses 
mow? 24 horses will mow more, therefore 24 is the 
second term in the proportion, and 8 is the first term. 

ThuS: 8: 24:: 128 

16:6 
24 X 6 X 128 18432 iAA . 
8X16 = T28~= 144 - Am ' 
Now take the next element in the proportion, which is 
days, and inquire how many acres can be mowed in 6 
days, if 128 acres can be mowed in 16 days; and as less 
can be mowed in 6 days than in 16 days, 6 is the second 
term of the proportion and consequently 16 is the first 
term, the terms being arranged in the proportion as 
if the answer depended upon days only, as shown above. 
Then divide the product of the second and third terms 
(24 X 6 X 128) by the product of the first terms (8 X 16), 
and the quotient is the answer, showing the number of 
acres that can be mowed by 24 horses in 6 days, if 128 
acres can be mowed in 16 days by 8 horses. 



130 A HAND BOOK FOR MECHANICS 

Then, in short, the rule to be followed when working 
out proportion by the process of compound proportion: 

First. Make that number which is like the answer 
(same denomination) always the third term. 

Second. Then take any two of the remaining numbers 
of the same denomination and consider whether the 
answer, if depending on them alone, would be greater or 
less than the third term, and arrange the numbers accord- 
ingly. 

Third. Then take any two of the other numbers of like 
denomination and arrange them as before, as if the answer 
depended upon them only, and so on until all are used. 

Fourth. Then divide the product of the second terms, 
multiplied by the third term by the product of the first 
terms, and the quotient is the required answer. 

Examples foe Exeecise 

1. If 15 men can build a house in 60 days by working 
15 hours per day. how long will it take 20 men to build 
the house if they work 10 hours per day ? 

20: 15:: 60 
10: 15:: 
15 X 15 X 60 13500 

20 X 10 ~ 200 " 62 " yS ' 

2. If 30 horses eat 12 tons of hay in 3 months, how 
much hay will 15 horses eat in 8 months? 

30: 15:: 12 
3:8 
15 X 8 X 12 1440 



30 X 3 90 



16 tons. 



3. If 10 men earn S120 in 7 days, how much will 7 
men earn in 4 days? Ans. S48. 



PROPORTION, OR THE RULE OF THREE 131 

4. If 15 men can plow 360 acres in 8 weeks, how many- 
acres can 27 men plow in 6 weeks? Arts. 486 acres. 

5. If 20 boys and 15 men can earn $300 in 12 weeks, 
how many dollars will 20 boys and 5 men earn in 9 
weeks? Arts. $75. 

6. If the interest on $200 is $10 for 12 months, what 
will be the interest on $1200 for 8 months? Arts. $40. 

7. If it takes $200 12 months to make $10, how long 
will it take $1200 to make $40? Ans. 8 months. 

8. How many dollars can 12 men earn in 14 days if 
7 men can earn $67 in 10 days? Ans. $160.8. 

9. If a man walks 200 miles in 30 days by walking 
9 hours per day, how many miles will he walk in 35 days 
if he walks 8 hours each day? Ans. 207H miles. 

10. How many days of 8 hours will it take a man to 
walk 207 miles if he can walk 200 miles in 30 days by 
walking 9 hours per day? Ans. 34 T 4 6 9 o days. 

INTEREST 

In order to perform examples coming under the head 
of this chapter, it will be necessary to know the names 
of the different terms used, and their meanings. 

By interest is generally understood the payment, or 
the sum paid for the use of money. 

If John hires from William $100 for one year, and 
pays William at the end of that time $105, he would be 
paying interest on the money lent him. 

That which is lent is called the principal, or base, 
and the sum paid for the use of it is called the interest, 
or percentage, and the specific rate charged for the 
use of the money is called the rate per cent ("per 
centum"). 

Percentage and per cent are both derived from the 
Latin words per and centum, which means by the hun- 



132 A HAND BOOK FOR MECHANICS 

dred or hundredths. Thus, 5 per cent means five one- 
hundredths (100) and 12 per cent means twelve one- 
hundredths (tVo). 

Then, if an article is bought for $100 and sold for $103, 
the gain is 3 per cent, because $3 is three one-hundredths 
of the original cost. 

Again, if an article is bought for $100 and sold for $97, 
the loss is 3 per cent. 

The sign % is used for the words per cent. Thus, 5% 
means 5 per cent, and 3% means 3 per cent. 
. Following are examples with rules for performing oper- 
ations in interest. The first set of examples shows how 
the percentage is obtained when the principal and rate 
per cent are given. The second set of examples shows 
how the rate per cent is obtained when the principal and 
percentage are given, and the third set of examples 
shows how the principal is found when the rate and 
percentage are given. 

Examples Showing how the Percentage is 
Obtained 

Example. What is 3% of $12? 

Process. Multiply the principal by the rate per 
cent. 

Thus: 





$12 




.03 




.36 cents 


Example. 


What is 5% of $75? 


Thus: 


$75 




.05 



$3.75 



INTEREST 

Example. What is 8% of 400 ? 

$400 
.08 



133 



$32.00 
Example. What is 7% of 126 yards? 

126 
.07 



8.82 yards 

Examples for Exercise 

1. What is 5% of $350? Ans. $17.50. 

2. What is 7% of $72? Ans. $5.04. 

3. What is 12% of $100? Ans. $12. 

4. What is 6% of 60 yards? Ans. 3.60 yards. 

5. What is 6% of 36 inches? Ans. 2.16 inches. 

6. What is 12% of 12 hours? Ans. 1.44 hours. 

7. What is 8% of $500? Ans. $40. 

8. What is 10% of $1000? Ans. $100. 



Examples Showing how the Rate Per Cent is 

Obtained when the Principal and 

Percentage are Given 

Example. What per cent of $36 is $9? 

Process. Divide the percentage by the principal, 
extending the division to hundredths. Thus: 



36)9.0(0.25 

72 

180 
180 



25% 



134 A HAND BOOK FOR MECHANICS 

Example. What per cent of 21 miles is 11? Thus: 

21)11. 0(0. 52ft = 522 8 T % 
105 

50 
42 

_8_ 
2 1 

Example. What per cent of $50 is $25? 

50)25.0(0.50 = 50% 
250 


Example. What per cent of $100 is $45? 

100)45.0(0.45 = 45% Ans. 
400 
500 
500 

Example. What per cent of 350 bushels is 75 bushels? 

Thus: 

350)75.0(0.211 = 21f% 
700 





500 




350 




150 15 3 




350 35 7 


Example. 


What per cent of $256 is $36? 




256)36.0(0.10A = 10A% 




256 




40 20 5 



256 128 32 



INTEREST 135 

Example. What per cent of $105 is $100? Thus: 

105)100.0(0.94?% Ans. 
945 

550 
420 
30 2 



105 7 

Examples for Exercise 

1. What per cent of $365.50 is 346? Ans. 94fff%, 

2. What per cent of $293 is 67? Ans. 22ff|%. 

3. What per cent of $76 is 39? Ans. 51 T 6 9%- 

4. What per cent of 367 is 38? Ans. 10Mt% 

5. What per cent of 245 is 103? Ans. 42 ? 2 9% 

6. What per cent of $5000 is $500? Ans. 10%. 

7. What per cent of 45 tons is 5 tons? Ans. 11|%. 

8. What per cent of $367.95 is $37? Ans. lOyW^o- 

Examples Showing how the Principal is Found, 

when the Rate Per Cent and Percentage 

are Given 

Example. $75 is 5% of how much money? 

Process. Divide the rate per cent into 100 and 
multiply the percentage by the quotient, and the product 
obtained is the required answer. 

Thus 5 )100 
20 
Then $75 
20 



$1500 Ans. 
Example. 12 is 40% of what number? 



136 A HAND BOOK FOR MECHANICS 

40)100(2^ 
80 



20 1 

40 — ^ 



Then 12 

z 2 





24 




6 
30 Ans, 


Example. 


40 is 12% of what n 




12)100(8J 
96 




_4 1 

12 — 3" 




Then 40 




8* 




320 




13i 



333? Ans. 



Examples for Exercise 

1. $2780 is 25% of how much? Ans. $11,120. 

2. $372 is 6% of how much money? Ans. $6,200. 

3. 79 yards is 7% of how many yards? Ans. 1.128f 
yards. 

4. 32 bushels is 7% of how many bushels? Ans. 
457} bushels. 

5. 28 is 5% of what number? Ans. 560. 

6. 100 is 10% o of what number? Ans. 1000. 

7. What number is 100 10% of? Ans. 1000. 

8. What number is 250 16% of? Ans. 1562J. 

9. 12 inches is 6% of how many inches? Ans. 200 
inches. 



INVOLUTION 



137 



INVOLUTION 

Involution is the raising a number (called the root) 
to any power. The powers of a number are its square, 
cube, fourth power, fifth power, etc. 

3X3 = 9. 9 is the square or second power of 3. 

3 X 3 X 3 = 27. 27 is the cube, or third power of 3. 

3 X 3 X 3 X 3 = 81. 81 is the fourth power of 3. 

Hence the second, third, fourth, fifth, sixth, etc., power 
is found by multiplying the number by itself 2 times, 
3 times, 4 times, as the case may be. 

The second power of a number is called its square, 
the third power .is called the cube, and the fourth, fifth 
sixth, etc., powers are called the fourth, fifth, sixth, 
etc., powers. Then, to square a number, multiply it by 
itself. 

Example. What is the square of 12 (written 12 2 )? 

12 
12 

144 Ans. 

And to cube a number, multiply the square of the 
number by itself again; that is, multiply the number by 
itself 3 times. 

Example. What is the cube of 12 (written 12 3 )? 

12 
12 



144 
12 



1728 Ans. 

Example. What is the sixth power of 12 (written 

12 6 )? 



138 A HAND BOOK FOR MECHANICS 



12 
12 



144 square 
12 



1728 cube 
12 



20736 fourth power 
12 



248832 fifth power 
12 



2985984 sixth power 

EVOLUTION 

Evolution is the process of finding the "root" when 
any power of a number is given, and is the reverse of 
Involution. 

There are only two cases where this is of much conse- 
quence: the finding the square root of a number and 
finding the cube root. 

Square Root 

To find the square root of a number is to find one of 
the two equal factors which produce it. 

Example. Find the square root of 589824. Thus: 

58,98,24(768 Ans. 
49 



146)998 
876 



1528)12224 
12224 



Process. Mark the figures off in two, from right to 
left, beginning, as we did in the above example, with the 



EVOLUTION 139 

last figure, 4; count two figures and mark the second, as 
shown, count two more and mark the figure, and so on 
till there are no more figures; then take the figure, or 
figures, to the left of the last dot, in this case 58, and find 
what number multiplied by itself will give 58. There is 
no number that will do so exactly, for 7 X 7 = 49 is 
too small, and 8 X 8 = 64 is too large. We take the 
one that is too small and place it in the quotient, and 
place its square, 49, under the 58, subtract and bring 
down the next two figures, 98. To get the divisor, 
multiply the quotient 7 by 2 = 14, place the 14 in the 
divisor and say 14 into 99 goes 6 times, place the 6 
after the 7 in the quotient and also after the 4 of the 
divisor; multiply the 146, now in the divisor, by the 6 in 
the quotient and place the product, 876, under the 998; 
subtract and bring down the next two figures, 24. To 
get the next divisor, multiply 76 by 2 = 152, see how 
often 152 goes into 1222, which is eight times, place the 
8 in the quotient after the 76 and in the divisor after 
the 2; multiply now the divisor 1528 by 8, and place 
the product under the 12224, then subtract. 

Example. Find the square root of 835. Thus: 

8,35(21.58 Arts. 
4 



41)435 
41 



425) 2500 
2125 



4308) 37500 
34464 

We proceed as before till we get to the remainder, 25, 
and we see it is not a perfect square; we wish the root to 
be taken to two or three places of decimals; there are 



140 A HAND BOOK FOR MECHANICS 

no figures to bring down, therefore bring down two 
cyphers and proceed as in first example; to the remainder 
attach two more cyphers and proceed as before, by 
attaching two cyphers to the remainder. In this way 
it may be carried to any number of decimal places. 

Example. Find the square root of 688.0625. 

Note this is a decimal quantity. In all such cases, 
instead of counting two from right to left, as in the 
preceding examples, we begin at the decimal point and 
count two toward the left and toward the right; we 
proceed then as in the other examples. Thus: 

6,88.06,25(26.23 Arts. 
4 



46)288 
276 



522) 1206 
1044 



5243) 16225 
15729 

Example. What is the square root of 17640.73205? 

ThuS: 1,76,40.73,20,5(132.818 Arts. 

1 



23) 


76 




69 


262) 


740 




524 


2648) 


21673 




21184 


26561) 


48920 




26561 



265628) 2235950 
2125024 



EVOLUTION 141 

Notice in this example the last figure is 5; always 
bring down two figures at a time, therefore bring down 50. 

Examples for Exercise 

1. Find the square root of 186624. Arts. 432. 

2. Find the square root of 3998.64. Ans. 63.234. 

3. Find the square root of 49.434961. Ans. 7.034. 

4. Find the square root of 588.0625. Ans. 24.25. 

5. Find the square root of 7986.57246. Ans. 89.367. 

CUBE ROOT 

To find the cube root of a number is to find one of 
the three equal factors which produce it. The process 
of extracting from a given number a certain specific 
number, which, when multiplied by itself three times, 
will equal the given number. 

Example. Extract the cube root of 46656. 

Process. Mark off three figures from the right, and 
there are 46 left. Find what number cubed will come 
to 46, or less than.it. Now 2 cubed = 8, that is (2 X 2 
X 2 = 8), 3 cubed = 27, 4 cubed = 64, so the number 
we need must be somewhat more than 3 but less than 4, 
we therefore take the 3. Arrange it as follows: 

46,656(36 Ans. 

_27 

96)2700)19656 
576 ) 
3276) 19656 

The three cubed is 27; put the 27 underneath the 46 
and subtract; the remainder is 19. Bring down the next 
three figures (656); multiply now the figures in the 
quotient by 3 and put the product well out to the left. 



142 



A HAND BOOK FOR MECHANICS 



Here it is 3 X 3 = 9. Put this 9 well out to the left, as 
shown; next multiply this 9 by the quotient figure 3, 
and place the result, 27, to the right of where the 9 is 
placed, and always put two cyphers after the product, as 
in example. 

This 2700 is called the " trial divisor." See how 
often it will go into 19656; it will go 7 times, but so 
close that we had better take 6. Put this 6 in the quo- 
tient after the 3 and also after the figure 9 on the left, 
making 96; then multiply the 96 by the new quotient 
figure 6 = 576, which place under the 2700; add them 
together and the sum, 3276, is the correct divisor; mul- 
tiply this correct divisor by the quotient figure 6 and 
place the product as shown; subtract, there being no 
remainder. 

The cube root of 46656 is found to be 36. 



Example. 
Thus: 

Column 
152 



1567 



Find the cube root of 146863664576. 



15816 



Column 2 

7500 

304 

7804 

811200 

10969 

822169 

83318700 

94896 



146,863,664,576(5276 
125 



)21863 



15608 



) 6255664 



5755183 



) 500481576 



83413596 ) 500481576 

Process. Mark them off in threes from right to left, 
and there are 146 to the left; 5, which placed in the 
quotient, is found to be the next less cube of 146, and 
the cube of 5 is 125; place the 125 under the 146 and 
subtract, and bring down the next three figures. 



CUBE ROOT 143 

Multiply the 5 by 3 = 15; put this well out to the left. 

Multiply the 15 by the 5 = 75. Put this down with 
two cyphers annexed = 7500, for the "first trial divisor;" 
it goes twice. Put the 2 in the quotient, also place it on 
the right of the 15 on the left; multiply 152 by 2 = 304, 
which place under the trial divisor, 7500; add them 
together and the sum 7804 is the "real divisor"; mul- 
tiply this by the quotient figure 2 = 15608, which put 
down and subtract, and on the right of the difference, 
6255, place the next three figures, 664. 

To find the third figure of the answer, multiply the 
52 in the quotient by 3 = 156. Place this well out to the 
left (column 1). Look at the middle column (column 2), 
and you will see the numbers 304 and 7804; add these 
together and to their sum add the square of the last 
figure in the quotient ( which is 2 in this case) and the 
square of 2 is 2 X 2 = 4. 

304 
7804 

4 

8112 

and this number, with two cyphers annexed, equals 
811200, is the new "trial divisor;" it goes 7 times. Put 
the 7 in the quotient and also after the 156 in column 1; 
now multiply 1567 by 7 = 10969, and put this under the 
trial divisor; add them together and the sum is the correct 
divisor, and is 822169, which multiplied by 7, and the 
product put down and subtracted, leaves 500481, on the 
right of which place the next three figures, 576. 

To find the fourth figure of the answer: Multiply the 
quotient 527 by 3 = 1581, which place in column 1, 
well to the left. Now look at the middle column (column 2) 
and add together the 10969, the 822169, and the square 



144 A HAND BOOK FOR MECHANICS 

of the last figure in the quotient in this case, the square 
of 7, equals 49. ^ 

822169 

49 

833187 

This number with two C3'phers after it is the new " trial 
divisor;" it goes 6 times. Put the 6 in the quotient and 
also after the 1581 in column 1; now multiply this 15816 
by the 6 in the quotient = 94896, which place under the 
trial divisor; add them together and multiply the sum 
by the 6. Put the product 500481576 down, subtract, — 
there is no remainder. Therefore, the cube root of the 
given number 146863664576 is 5276. 

Following is another method of extracting the cube 
root which may, perhaps, be considered simpler. 
Find the cube root of 1728. 

1,728(12 Arts. 

I 

3 x (10) 2 = 300) 728 
3 X 10 X 2 = 60 
2 2 = 4 

364) 728 

The advantage of this method is that when the pupil 
has mastered the workings of the first figure of the 
quotient, the rest of the sum is at his command, as the 
process is the same for any number of figures, the same 
for twenty as for two. 

Process Explaining how Examples are PERFORMEd 
by the Above Method 

The number 1728 was first marked off into quantities 
of three figures as before, and selecting the next lowest 



CUBE ROOT 145 

cube for the first period, in this case 1, which place in 
quotient and divisor, l 3 = 1, which place under the 1 in 
the dividend. Now bring down next three figures (728). 
Take now the number in quotient (in this case 1) add a 
nought to it, making it 10; this always square and mul- 
tiply by the constant 3. Thus 3 X (10 2 ) = 300; this 
forms a trial divisor. For the next line, still use the 3 
which now multiply by 10 (not 10 2 this time), multiply 
product thus obtained by that figure in quotient, ob- 
tained from the trial divisor; thus 3 X 10 X 2 = 60, 
which place under the 300. For the third line simply 
square the number obtained by the trial divisor in this 
case 2 2 = 4, which place under the 60 then add all 
quotients together. 

3 x 10 2 = 300 

3 X 10 X 2 = 60 

2 2 = 4 

364 = divisor. 

Example. Find the cube root of 2438569736. 

2,438,569,736(1346 Arts. 
1 



3 x 10 2 = 300)1438 

3 x 10 X 3 = 90 

3 2 = _9 

399)1197 



3 x (130 2 ) = 50700) 241569 
3 X 130 X 4 = 1560 

4 2 = 16 

52276) 209104 
3 X (1340 2 ) = 5386800) 32465736 
3 X 1340 X 3 = 24120 

6 2 = 36 

5410956) 32465736 



146 A HAND BOOK FOR MECHANICS 

Examples for Exercise 

Find the cube root of the following numbers: 

1. 16934.994432. Ans. 25.68. 

2. 134217728. Ans. 512. 

3. 80677568161. Ans. 4321. 

4. 15069223. Ans. 247. 



PART II 

ARITHMETICAL SIGNS AND CHARACTERS AND 
EXPLANATION OF FORMULA 



ARITHMETICAL SIGNS 

By a knowledge of the meaning of the different arith- 
metical signs and by the proper use of same, we are 
enabled to perform arithmetical problems in a brief way. 

The sign + reads "plus/' and "more," and means 
that the number after it is to be added to the number 
before it; thus, 3 + 5 are 8. 

The sign — reads "minus/' "less," signifying subtract, 
and means that the number after it is to be subtracted 
from the number before it; thus, 5 — 3 leaves 2. 

The sign X reads "multiply by," and means the 
number before it is multiplied by the number after it; 
thus, 2 X 2 are 4. 

The sign -f- reads "divide by," and means that the 
number before it is to be divided by the number after it; 
thus, 4 + 2 is 2. 

The sign : reads "divide by," also, and means that 
the number after it is to be divided by the number 
before it; thus, 4: 2 is 2. 

The sign V reads divided by 3, and is the same thing 
as 18 -7-3 equals 6. 

The sign = reads "equality," or "equal to," and means 
that the quantity after it is equal to or of the same 
value as the quantity before it; thus 3 + 6 = 9, or 
6 - 3 = 3, or 6 + 2 X 2 = 16. 

The sign 5 2 reads 5 squared, and means that 5 is to 
be multiplied by itself; thus, 5 X 5 = 25; then 25 is 
called the square of 5. And 5 3 reads, 5 cubed, 5X5x5 
= 125, and 5 4 means that 5 is to be multiplied by itself 
four times. 

149 



150 A HAND BOOK FOR MECHANICS 

The sign V reads the " square root," and means that 
the square root of the number alongside it is to be 
considered; thus, \/36 reads the square root of 36 and 
means that number which, when multiplied by itself, 
gives 36; thus, 6 X 6 = 36, hence 6 is called the square 
root of 36. Sometimes the sign of square root is rep- 
resented thus, -\/36, which reads what number, when 
squared, is equal to the number 36. 

The sign fy reads the "cube root" and means that the 
cube root of the number alongside it is to be considered; 
thus, ^/216 reads the cube root of 216 and means that 
number which, when multiplied by itself three times, 
gives 216; thus, 6X6x6 = 216, hence 6 is the cube 
root of 216. 

Example. 5 X ^/216 is what? The question here is 
what is the product of 5 multiplied by the cube root of 
216, and as the cube root of 216 is 6, then 5 X 6 = 30. Ans. 

The sign ~ reads the difference between and means 
that the less number, whether before or after it, is to be 
subtracted from the greater number; thus, 5 ~ 2 = 3, 
and 2 — 5 = 3. 

The sign () is called "brackets." Brackets are of 
various shapes, thus: (), {}, []. 

Where brackets are used it is meant that all the quan- 
tities within each set of brackets are to be put together 
first and treated as a whole; thus (12 + 4 — 3) — (12 + 
3 — 7) means that the 4 is to be added to 12 = 16, and 
that 3 is to be subtracted; therefore, 16 — 3 = 13, hence 
the quantities in the first set of brackets, when put 
together, become 13. 

Take now the quantities between the next set of 
brackets, which treat in the same way; thus, 12 + 3 = 
15 — 7 = 8. Here we have 8 as a result of putting the 



ARITHMETICAL SIGNS 151 

quantities together as directed by the signs. The sum 
is now 13 — 8 = 5. Arts. 

Sometimes, instead of brackets, a line called a vinculum 
is drawn over the expression which is to be treated as a 
whole. Thus, 6X4 + 3x2 + 1 means that the pro- 
duct of 6 X 4 = 24 is to be added to the product of 
3x2 = 6, and to the sum of the products is to be added 1. 
Thus: 24 

6 
_1 

31 Ans. 



Thus, in the above example, 6x4 + 3x2 + 1 is 

equivalent to (6 X 4) + (3 X 2) + 1. 

The sign •.■ reads " because," or "since." 

The sign .-. reads "therefore," or "hence." 

The sign > indicates that the number which is before 

it is greater than the number which follows it. 

The sign < indicates that the number before it is 

less than the number which follows it. 

The sign ^ reads, or stands, for the words "is not 

equal to." 

The sign < stands for the words "is not less than." 
The sign > stands for the words "is not greater than." 

Examples on the Use of the Addition Sign 

The addition sign requires little notice, because when 
placed before a number it indicates that it is to be 
"added" to what has gone before. Thus, 3 + 9 means 
that 9 is to be added to 3, and 3 + 9 + 2 means that 9 
is to be added to 3 and then 2 added to the result. Thus, 
in addition, the order of performing operations is from 
left to right. 



152 A HAND BOOK FOR MECHANICS 

Examples for Exercise 
What is the sum of 2 + 6 + 9? Arts. 17. 
What is the sum of 15 + 3 + 1? Arts. 19. 
What is the sum of 7 + 9 + 6 + 8? Arts. 30. 
What is the sum of 8 + 2 + 3 + 1? Arts. 14. 
What is the sum of 6 + + 3 + 5 + 7? Ans. 21. 

Examples on the Use of the Minus Sign 

This sign requires little notice also, because when 
placed before a number it indicates that it is to be sub- 
tracted from what has gone before. Thus, 4 — 2 means 
that 2 is to be subtracted from 4, and 16 — 4 — 3 means 
that 4 is to be subtracted from 16 and then 3 is to be 
subtracted from the result. Thus, 16 — 4 = 12, and 
12 - 3 = 9. Ans. 

Then in subtraction, the order of performing opera- 
tions is from left to right, as it was in addition. 

Examples for Exercise 

What is the result of 25 - 3 - 4? Ans. 18. 
What is the result of 19 - 2 - 10? Ans. 7. 
What is the result of 16 - 8 - 8? Ans. 0. 
What is the result of 100 - 50 - 25? Ans. 25. 
What is the result of 300 - 50 - 100 - 100? Ans. 50. 

Examples Performed where the Addition and 
Minus Signs are Involved 

Note when the first term of any quantity has no 
sign before it the + (plus) sign is always understood to 
be there. 

Example. What is the sum of 12 + 4 - 2 + 7 - 4. 

Here we add together all those terms which have the + 
sign actually before them as +4 + 7, and the term 



ARITHMETICAL SIGNS 153 

which is understood to have the + sign before it, al- 
though the sign is not put down; in this case it is the 12. 
Then 12, 4, 7 added together make 23, and — 2 and — 4 
added together make —6. The sum then becomes 
23 - 6 = 17. Ans. 



Example. 


Find 


the sum of 24 - 16 + 4 - 


- 7 + 




Thus 


+ 24 + 4 + 4 = 


+ 32 






And 


- 16 and - 7 = 


- 23 

9 


Ans. 


Example. 


What does the following 


come to? 






-6 + 3 + 2-5 







Proceed as before by adding together all those terms 
which actually have the + sign before them, and then 
add together all those signs having the — sign; sub- 
tract, then, the smaller from the larger, but put before 
the remainder the same sign as that of the larger 
quantity. The above example, then, will be solved thus: 

+ 3 + 2 = 5 sum of terms having the + sign 
— 6 — 5 = — 11 sum of terms having the — sign 
Then - 11 

+ A 

— 6 Ans. 

Example. -8 + 12 + 2-9 + 3-8 + 2-17 
comes to what? 

Thus - 8 +12 

- 9 +2 

- 8 +3 

- 17 + _2 

- 42 +19 
Then - 42 

+ 19 

- 23 Ans, 



154 A HAND BOOK FOR MECHANICS 

Note that the answers of both above examples have 
the — sign before them, because the sum of the quan- 
tities after the — signs are the larger. 



Examples for Exercise 
What do the following come to? 

1. 8 — 12-2 + 9-3 + 8-2 + 17. Ans. + 23. 

2. -4 + 3-7-8 + 2. Ans. - 14. 

3. 5-17 + 3 + 2. Ans. - 7. 

4. 25-6-2 + 7. Ans. + 24. 

5. 10 - 25 + 3 + 15 - 17. Ans. - 14. 

Examples Showing how Brackets ( ) are Used 

Example. -3 + 4 + 6- (5 + 6) +8- (9-3) + 
(14 — 4 — 4) is equal to what? 

Process. Get rid of all the brackets first by putting 
the quantities within each set of brackets together so 
that they become a single number. Then to solve the 
above problem we proceed to deal with the terms inside 
each bracket first. 
Thus 

(5 + 6) = 11 (9 - 3) = 6 (14 - 4 - 4) = 6 

Hence 

- 3 + 4 + 6 - (5 + 6) + 8 - (9 - 3) + (14 - 4 - 4) 

Becomes 

-3 + 4 + 6-11+8-6 + 6 

Add now all the + signs together and all the — signs 
together, subtract the larger from the smaller and the 
result is the required answer. 



ARITHMETICAL SIGNS 155 

Thus +4 - 3 

+ 6 - 11 

+ 8 - _6 

+ _6 - 20 

+ 24 
- 20 
+ 4 Ans. 

Example. 7-8+6- (5-6 + 3) +9+ (9 -3 + 2 
— 6 + 7) is equal to what? 

Thus 

7-8 + 6- (5-6 + 3) +9+ (9-3 + 2-6 + 7) 

Equals 

7-8+6-2+9+9 

Adding all the + signs together we get 31 
Adding all the — signs together we get 10 

Thus 31 - 10 = 21 Ans. 

Note. When no sign is between the quantity outside 
the bracket and the bracket, it means that the quantity 
within the bracket is to be multiplied by the quantity 
outside. 

Example. 3(4 - 2 + 3) - 2 + 6 is equal to what? 

Here outside the brackets is 3, having no sign between 
it and the bracket, which means that after reducing the 
numbers within the brackets to a single number we 
multiply it by 3. 

Thus 3 (4 - 2 + 3) = 5 multiplied by 3 = 15 
Hence 3 (4-2 + 3) -2 + 6 
Equals 15-2 + 6 

Adding all the plus quantities together, and all the 
minus ones separately, we get, 21 — 2 = 19 Ans. 



156 A HAND BOOK FOR MECHANICS 

Example. 4 (3 + 2 - 6 + 4) + 2 (3 + 6 - 2) is equal 
to what? 

Thus '4 (3 + 2-6 + 4) +2 (3 + 6- 2) 
Becomes 3x4+7x2 

Which is 12 + 14 

Which added together is: 

12 

14 

26 Arts. 

Examples for Exercise 

The following equals what? 

1. - 2 + 12 + 3 - (9 + 4) + 3 - (7 - 5) + (6 - 2 
+ 7). Arts. 12. 

2. 4 (3 + 2 - 6 + 4) + 2 (3 + 6 - 2). Ans. 26. 

3. 4 (3 + 2 - 6 + 4) - 2 (3 + 6 - 2). Ans. - 2 

4. 3 (4 - 2 + 3) + 2 - 6. Ans. 11. 

5. - 2 + 6 + 2 - (4 + 3) + 4 - (7 - 5) + (7 - 1 
+ 2). Ans. 9. 

Examples on the Proper Use of the 
Multiplication Sign 

The multiplication sign, X, is placed between two 
numbers to indicate that the first number is to be mul- 
tiplied by the second. Thus, 12 X 6 means that 12 is to 
be multiplied by 6; also 12 X 6 X 3 means that 12 is 
to be multiplied by 6 and the result multiplied by 3. 
Sometimes the sign X is replaced by a point or dot 
which is placed in a line with the bottom of the quan- 
tities to be multiplied. Thus, 12 X 6 X 3, and 12.6.3, 
mean the same thing, namely, that 12 is to be multiplied 
by 6 and the result by 3. 

Because of the improper use of the multiplication sign 



ARITHMETICAL SIGNS 157 

it causes more trouble than any of the other signs. For 
instance, in a quantity such as follows 7 + 2x4 we 
would be apt to say 7 and 2 are 9, which, multiplied by 
4 = 36. Now this would be wrong. The first step 
would be to multiply the 2 by 4 and then to the result 
add the 7. Thug 7 + 2 x 4 

Becomes 7 + 8 
Equals 15 Arts. 

We learn, then, from the above, that the first step to 
take when the multiplication sign is involved, is to get 
rid of it in the manner shown. That is, always get rid 
of the multiplication sign first, unless brackets are used, 
in which case, we would get rid of all brackets first, and 
then dispose of the multiplication signs. 

Example. 4-3X2 + 6X4-2X3 + 8X2 is 
equal to what? 

Thus 4-3X2 + 6X4-2X3 + 8X2 

Becomes 4-6 + 24-6 + 16 

Now put the terms together as in addition and sub- 
traction, and we have 44 — 12 = 32 Arts. 

Examples. What does the following equal? 





-3 + 4 + 2X6X2- 


3 + 4X6 


Thus 


-3 + 4 + 2X6X2- 


3 + 4X6 


Becomes 


-3 + 4 + 24-3 + 24 
+ 4 + 24 + 24 = + 52 




And 


-3+-3 --6 






46 


Arts. 



Example. What is the following equal to? 

24 - (32 - 15) + 3 X (9 - 5) 

Process. Get rid of all the brackets and then get rid 
of the multiplication signs. 



158 A HAND BOOK FOR MECHANICS 

The example, 24 - (32 - 15) + 3 X (9 - 5), becomes 
24 — 17 % + 3 X 4. The brackets are here disposed of. 

Now dispose of the multiplication sign and we have 
24 - 17 + 12. 

Proceed now as in addition and subtraction. 

24 
12 
36 - 17 = 19 Arts. 

Examples for Exercise 
Find the value of the following: 

1. 6 + 4 X (9 - 3) + 7 - 2 X (45 - 41). Arts. — 41. ' 

2. 54 - (64 - 29) + 6 X (18 - 5). Arts. 97. 

3. 24 - (32 - 15) + 3 X (9 - 5). Ans. 19. 

4. -3 + 4 + 2X6X2-3 + 4X6. Ans. 46. 

5. 4-3X2 + 6X4-2X3 + 8X2. Ans. 32. 

Examples on the Use of the Division Sign 

Example. 5 + 2X(7-2)-4-8-h(3 + 5) equals 
what number? 

Process. First get rid of all the brackets; secondly, 
get rid of the multiplication and division signs, then 
proceed as in addition and subtraction. 

Example. 5 + 2 X (7 - 2) - 4 - 8 -r- (3 + 5). 

Brackets left out, 5 + 2X5-4-8^8. 
Multiplication and division signs done away with. 
5 + 10-4-1. 

+ 5 + 10 = +15 
-4 + - 1 = - _5 

10 Ans. 
Example. 32 + 4 + 2. 



ARITHMETICAL SIGNS 



159 



Example. 



Thus 4)32 
2)8 
4 
32-4 + 2. 

Thus 4)32 



Ans. 



8 
_2 

10 Ans. 

Example. What is 24 - 6 + 4 — 2 equal to? 

Thus 6)24 
4 
4 
8-2 = 6 Ans. 

Example. What does the following amount to? 

(3 + 5) X 6 - 2 X (2 + 2) - 4 - 5 + 10 
Thus 8X6-2X4-4-5 + 10 
+ 48 - 2 

+ 10 - 5 

+ 58 - 7 = 51 Ans. 

Example. 2X4—2 amounts to what? 

8^-2 = 4 Ans. 

Sometimes division is represented thus: 

4 + 2 
6-4 

where one quantity is placed above another with a line 
between them. When thus represented the value of the 
top quantity is to be divided by the value of the bottom 
quantity. That is, in the above example the 4 + 2 is to 
be divided by the 6 — 4. 



160 



A HAND BOOK FOR MECHANICS 



Thus 
And 6 



- 2= 6 
2 



Example. 



- 4 

Then 6-2 = 3 Ans. 
5 + 4 X 6 is equal to how many? 
2 + 3X6 

Thus 5+4X6 = 29 
And 



Example. 



3 



2 + 3 X 6 = 20 
Thus 29 ~- 20 = lft 

12 - 4 + (6 - 4) 



Ans. 



(6 - 4) + (16 - 8) 
Such examples as the above sometimes cause trouble. 
Note that the 3 before the fraction is placed on the 
level of the division line, and in this position it indicates 
that the whole fraction is to be multiplied by it. It 
will be remembered, however, when multiplying a frac- 
tion by a whole number, that the numerator of the 
fraction is multiplied only. Then in the above example 
the top line only is multiplied by the 3. 

12 - 4 + (6 -- 4) - 8 + 2 = 10 

10 



Thus: 3 



Then 3 



+ 8 = 
= 3 Ans. 



(6 - 4) + (16 - 8) = 2 
10 30 
10 = 10 

The pupil would better now refer to vulgar fractions 
and become very familiar with the fact as there explained, 
that when multiplying a fraction by a whole number the 
numerator of the fraction is multiplied only. 

Example. 3 (12 - 4 + 6 - 4) - (6 - 4 + 16 - 8) is 
equal to how many ? 

Thus 3 (12 - 4 + 6 - 4) - (6 - 4 + 16 - 8) 

Becomes 10 -r- 10 

3^ 

30 -=- 10 = 3 Ans. 



ARITHMETICAL SIGNS 161 

It will be remembered, when no sign is placed between 
a quantity and a bracket, that the quantity within the 
bracket is to be multiplied by the quantity without. 
Therefore, after reducing the quantity in the first set 
of brackets to 10, we multiplied by 3 and then divide 
by the sum of the quantities within the next brackets. 

The following example will impress this more fully 
upon the mind: 

Example. Find the value of 4 + 2 (2 + 6) - 3 + 2 
(9 - 4), which is the same as4 + 2X(2 + 6)-3 + 2 
X (9 - 4). 

Hence 4 + 2X8-3 + 2X5 
4+16-3+10 

4 
16 
10 
30 - 3 = 27 Ans. 

Examples for Exercise 

Find the value of: 
L 7 + 7 - (6 - 2 ) 

(5 - 2) + 3 - 1 AnS ' *' 

2. 14 + 6 - (2 + 4) -r (6 + 8 - 12). Ans. 17. 

3. 9 + 4 X (7 - 3) - 8 - 16 -T- (9 - 5). Ans. 13. 

4. 12 - 4 + (6 - 4) 



(6 - 4) + (16 - 8) 



Ans. 3. 



Exercises on the Signs Representing the Power 
of Numbers, as 4 2 , 4 3 , 4 4 , etc. 

The powers of a number are its square, cube, fourth 
power, fifth power, etc. 



162 A HAND BOOK FOR MECHANICS 

Thus 4 2 is equal to 4 X 4 = 16 

4 3 is equal to 4 X 4 X 4 = 64 

4 4 is equal to 4 X 4 X 4 X 4 = 256 

Especial names are given to 4 2 and 4 3 ; they are called 
respectively, the square and cube of 4. 

The small figure, or letter, placed above a quantity, to 
indicate the number of times that quantity is to be taken 
as a factor is called the " Index," or " Exponent." 

Thus, 4 2 means that the factor 4 is to be taken 2 times 
and 2 is called the index. Therefore, always multiply 
the given number that number of times by itself, as 
indicated by the " Index," and the continued product 
is the required power. That is, 4 to the eighth power 
(4 8 ) is 4X4X4X4X4X4X4X4 = 65538 

Example. 4 2 + 6 3 is equal to what? 

4 2 = 16 
6 3 = 216 

232 Ans. 

Example. 4 3 — 3 3 is equal to what? 

4 3 = 64 

3 3 = 27 Then 64 - 27 = 37 Ans. 

Exercises on the Sign that Represents the Roots 
of Numbers a/ and ^/ and \/~ 

A number which, when squared, is equal to some 
specific number is called a " square root," and is repre- 
sented by the symbol y/ and sometimes \V. Thus 6 is 
\/ 36, since 6 2 = 36, 

And a number which, when cubed, is equal to any 
specific number is called a "cube root," and is rep- 
resented by the symbol ^/ . Thus 4 is ^/ 64, since 
4 3 = 64. 



ARITHMETICAL SIGNS 163 

The sign (>/), which represents the root of numbers, 
was originally the first letter of the word radix, which 
word really means the source from which anything 
springs. The sign is now called the " radical sign/' and 
the radical sign is common to all roots. Thus, when it 
is required to express the square root of a number or 
quantity, we simply place this sign before it, as V 144, 
and when it is required to express the cube root of a 
number or quantity, the same sign is placed before the 
number with a 3 in the elbow, thus %/. 

Sometimes a number composed of two or more terms 
is to have its root expressed. In all such cases, place 
the radical sign in front and draw a line over the num- 
bers whose root is required as far as they extend. 

For example, \/3(4 + 2 + 3) means that the square 
root of the sum of those numbers under the line is to be 



expressed, also \/2 + 14 means that the square of the 
sum of the numbers underneath the line is required. 

Also ^/2 + 6 expresses that the cube root of the sum 

of the numbers under the line is required. 

Following is another way, which is not often used, for 

expressing that the root is required. 
i 
Thus, 16 2 means that the square root of 16 is required. 
i 
And (2 + 14) 2 means that the square root of the sum 

of the numbers within the brackets is required. Some- 

3. 

times the power and root are combined, as 4\ This is 
read as the square root of 4 cubed. In all such cases 
the numerator figure represents the power, and the 
denominator figure represents the root. In the above 
example 4 cubed = 64 and the square root of 64 = 8. 



164 A HAND BOOK FOR MECHANICS 

Law of Signs in Multiplication 

For those who are not acquainted with algebra it is 
difficult and very confusing to know how to determine 
the sign of the product. The rule by which the sign 
of the product is determined is called the "Law of 
Signs/' 

Rule 1. Plus and minus multiplied together always 
give minus. 

Rule 2. Minus and minus multiplied together always 
give plus. 

In other words, when multiplied together like signs 
give + and unlike signs give — . For example: 

-3 X + 2 = -6 

Here minus 3 is multiplied by plus 2 and the sign of 
the product 6 is a minus sign, thus — 6, product. 
Because unlike signs multiplied together give — • 
(Rule 1.) 

Example. 8 + 4-3X(3-2 + 4)is equal to how 
much? 

First gather together the numbers inside the brackets 
and we have 8 + 4 _ 3 x (3 _ 2 + 4) 

Becomes 8 + 4 - 3 X +5 

When a number has no sign before it, remember that 
the + sign is always understood to be before it, hence 
the 3 in the brackets is + 3, which, when added to the 
-f 4 = +7, from which — 2 is subtracted, and we 
have + 5 left. Therefore, when the numbers inside the 
brackets are gathered up we have + 5, which is to be 
multiplied by — 3. The sum then is: 

8+4-3X +5 

8 + 4-15= -3 Ans. 



ARITHMETICAL SIGNS 165 

Example. What is the value of 

8 + 14 - 6 X (10 - 14 + 6) 
Becomes 8 + 14-6X+2 
Becomes 8+14-12 
Becomes 8 + 14 = 22 - 12 = 10 Arts. 

Example. What is the product of — 6 X — 6? 

Thus -6X-6=+36 Am. 

Here minus 6 is multiplied by minus 6 and the sign of 
the product is plus. Because like signs multiplied 
together give + (Rule 2). 

Example. What is the product of + 6 X +6? 

+ 6 X + 6 = + 36. Am. 

Example. 13 - 9 X 4 - 11 X (5 - 42 + 29) equals 

What? 13 -36- 11 X -8 

This is an example of Rule 2, being — 11 X — 8 = 
+ 88. 

Hence 13 - 36 + 88 = 65 Am. 

Example. What is. the value of 

32 - 8 - 11 - 9 X (32 - 71 + 29) 
4 - 11 - 9 X - 10 

This again is an example of Rule 2, being minus 9 X 
minus 10. Thus - 9 X - 10 =+ 90. (Like signs 
give + .) 

Hence 4 - 11 + 90 = 83 Am. 

The + sign is not placed before the answer, 83, of the 
above example, because when no sign is before a quantity 
+ is always understood to be there. 



166 A HAND BOOK FOR MECHANICS 

Law of Signs in Division 

In division we have the same "Law of Signs" as in 
multiplication. 

Unlike signs give — . (Rule 1.) 

And like signs give + . (Rule 2.) 

That is, if the signs of the dividend and divisor are 
unlike, the sign of the quotient is minus ( — ), and if the 
signs of the dividend and divisor are like signs, the sign 
of the quotient is plus ( + ). 

For example: — 12 -= 4 = +3 Ans. 

The answer here is plus 3 because the sign of the 
dividend (-12) and the sign of the divisor (- 4) are 
like signs, and like signs give + . (Rule 2.) 

Example. + 12-+4=+3 Ans. (Rule 2.) 

Example. + 15 -s- — 3 = - 5 Ans. (Rule 1.) 

In the above example the quotient sign is minus, 
because the sign of the dividend and the sign of the 
divisor are unlike. 

Example. -16-h+4=-4 Ans. (Rule 1.) 

Example. (9 X 24 + 6) -(4 + 2x8- 131). 

216 + 6 -4 + 16 - 131 
+ 222 -?■ - 111 

This is now an example of Rule 1. 

+ 222 h 111 = - 2 Ans. 

Miscellaneous Examples for Exercise 

1. 42-7 + 6-2. Ans. 10. 

2. 56 4- 4 - 3 X 6 + 4. Ans. 0. 

3. 250 -f- 10 + 4 - 2 X (26 - 13). Ans. 3. 

4. (52 - 8.14) X (8 + 4). Ans. 526.32. 

5. (52 - 8.14) X 8 + 4. Ans. 350.92. 



ARITHMETICAL SIGNS 167 

6. ( 42.4 - 12) 2 - 8 X 4. Arts. 21.76. 

64 - 23 

7. What is the value of (12 + 3) 3 - (21 - 4) 2 . Ans. 
. 3014. 

8. (4.05) 2 . Ans. 8398.08. 

(i) 3 

9. (6.012 + 0.050) X (.070 + |). Ans. 3.395. 

10. (2.562 + 6.0002) - (0.0642 + 2.9808). Ans. 5.5172. 

11. 3 + 4 - 2 X 4 - 3 X (8 - 28 + 6). Ans. 41. 

12. 28 -^ 4 - 6 - 12 X (- 8 + 30 - 10). Ans. - 143. 

13. 18 - 7 X 2 + 9 -T- 3 - (16 - 51 + 13). Ans. 15. 

14. What is the value of 29.4 2 - \/25? Ans. 107420. 

.008 



Formula 

A formula is a rule expressed in a brief and concise 
way, by the means of certain letters and arithmetical 
signs. The pupil, therefore, in order to read or solve a 
formula clearly and easily must be thoroughly acquainted 
with the arithmetical signs, and the operations which they 
indicate are to be performed. 

The reading and solving of formula may appear to 
those who have not had the early school advantages 
very difficult; however, for such, if the following rules 
which govern the reading and solving of formula gener- 
ally are given careful study and thought, many of the 
problems which the mechanic comes across in his daily 
walks of life, and which have no earthly meaning to him, 
will be clearly read and easily solved. Formulae are 
used for expressing general rules in mathematics and 
physics. 

For instance, the rule of estimating the nominal horse- 
power for ordinary condensing marine engines is the 



168 A HAND BOOK FOR MECHANICS 

square of diameter of cylinder in inches multiplied by 
the number of cylinders and the product divided by 30. 
This rule may be expressed by the following formula: 

Example 1. p2 x N _ Am 

30 

D = Diameter of cylinder in inches. 

N = Number of cylinders. 

So it is seen by means of two or three letters and signs 
we can express a long rule. This fully explains the ob- 
ject of all formulae. Again, the rule for finding the nomi- 
nal horse-power of paddle steamers is to square the 
diameter of the piston and multiply by its velocity, and 
the product divided by 6000 gives the nominal horse- 
power. Now this rule may be expressed by the formula: 

D 2 X V = nominal horse-power 
Example 2. — 

Suppose, in the above example, the diameter of the 
piston is 24 inches, stroke 2 feet, revolution 44 per 
minute, what would the nominal horse-power be? 

Thus 24 2 X 176 = 15.981 nominal horse-power. 
" 6000 

The above examples are given only to demonstrate 
that a formula which is composed of a few letters and 
signs will express a long rule, and the student is not 
supposed to try to solve same until he at least has become 
familiar with the following examples, which explain the 
solving of formulae. 

Examples Showing how Formulae are solved 

By referring to the above examples, 1 and 2, it will be 
seen that in a formula letters represent a value or some 



ARITHMETICAL SIGNS 169 

dimension. Thus D 2 in example 1 represents a dimen- 
sion and N represents a number or quantity. And in 
example 2 it is seen that D 2 represents 24 2 and V repre- 
sents a number equal to 176. 

The first step to take, then, in solving a formula is to 
substitute figures for letters, then proceed as directed by 
the arithmetical signs of the formula. 

The following examples will clearly demonstrate the 
process of solving all formula?. 

Example. IfA=B + C-D + E-F, what must 
the value of A be when B = 12, C = 8, D = 5, E = 9, 
and F = 10? 

We first proceed by substituting the figures for the 
letters, thus: 

A = 12 + 8-5 + 9- 10 

Then proceed as directed by the signs. 

A = 29 - 15 

= 14 Ans. 

Example. IfX = A + B-C + D-F, what is the 
value of X when A = 20, B = 14, C = 9, D = 8, and 
F = 12? 

First substitute figures for the letters, thus: 

X = 20 + 14 - 9 + 8 - 12 

Then proceed as in the arithmetical part. 

X =42-21 
= 21 Ans. 

Example. If K = h X - i D + J C - f F, what 
is the value of K when A = 12, D = 24, C = 30, and 
F = 12? 



170. A HAND BOOK FOR MECHANICS 

As A = 12, then § A = 6 

And as D = 24, then \ D = 6 

And as C = 30, then \ C = 6 

And as F = 12, then f F = 9 

Then K = 6-6 + 6-9 

=12-15 

= — 3 Arts. 

Example. IfK = 3A + 4B + 9C-7D-E + 3F, 
find the value of K when A = 12, B = 5, C = 2, D = 4, 
E = 12, F = 1. 

Here 3 A = 3 times 12 = 36; 4 B = 4 times 5 = 20; 
9 C = 9 times 2 = 18; 7 D = 7 times 4 = 28; E = 12, and 
3 F = 3 times 1 = 3. 

Hence K =36 + 20 + 18-28-12 + 3 
= 77-40 
= 37 Arts. 

Note when two or more letters are together, without 
any sign between them, it is always understood that they 
are to be multiplied together. 

Thus D N is the same as D X N 
A~B A X B 

Example. What is the value of N when N = AB + 
CD - EF; when A = 4, B = 5,C = 3,D = 4, E = 2, F = 10? 

N=4X5 + 3X4-2X10 
= 20 + 12-20 
= 32-20 
= 12 Ans. 

Example. If M = AB - CD + EF - G, what is the. 
value of M when A = 2, B = 3, C = 4, D = 5, E = 6, 
F = 7, and G = 20? 



ARITHMETICAL SIGNS 171 

M=2x3-4x5 + 6x7-20 
= 6-20 + 42-20 
= 48-40 
= 8 Arts. 

Example. If A = C — f C j, find the value of A 

when C = 8, D = 20, C = 10. 

A = 8 - (20 - 10) 
4 

= 8-5-10 
= 8-15 
= — 7 Arts. 

Example. If N = C - (- - pY find the value of N 
when C = 8, S = 3J, P = 1|. 



N = 8 - (3i - 1J) 
2 
= 8 - (If - 1§) 
= 8-i 
= 7J Ans. 

Example. If M = 4 ABC - 3 ED - 5 EFG, what is 
the value of M when A = 2, B = 3, C = 4, D = 5, 
E = 6, F = 4, G = 2? 

M=4X2X3X4-3X4X 5- 5X6X4X2 
= 96 - 60 - 240 
= 96 - 300 
= - 204 Ans. 

Note. In the above example there is no sign between 
the figures and the letters. 

The sign X is generally omitted between letters or be- 



172 A HAND BOOK FOR MECHANICS 

tween a figure and a letter. Thus AB means the same 
as A X B and 4 AB the same as 4 X A X B. 

Example. If N = A B , what is the value of N, 

C - D 
when A = 6, B = 7, C = 16, D = 10? 

N = 6 X 7 = 42 



16-10 6 



= 7 Ans. 



B C 
Example. If M = A^ — rr what is the value of M 

D — E 

when A = 2, B = 3, C = 4, D = 16, and E = 8? 

„ n 3X4 2X3X4 24 

M=2 T6~8 = —8 = ¥ = 3 Am ' 

Example. What is the value of 2 H — — — when 

A = 82 and a = 38? 

82 + 38 -64 _jg_ = 2 + .056 = 2.028 Ans. 

2000 2000 

A "D _i_ p 

Example. What is the value of 2 -± when 

A = 20, B = 6, C = 16? 

20-6+16 . 30 _ ^ 30 60 

2 = 2 — = 2X- = -t = 4 4ns. 

15 15 15 15 

D 3 — d 3 

Example. 6 : — W T hat is the value of this when 

A 

D = 14, d = 12, and A = 15? 

14 3 - 12 3 2744 - 1728 n 1016 6096 

6 — IT" =6 h5 6X -l5- = -l5~ 

= 406.4 Am. 

Example. A - (B ~ § }\~ K) X .003 N. 

What is the value of the above when B = 120, C = 32, 
T = 67, K = 32, and N = 3375, M = 120, R = 67? 



ARITHMETICAL SIGNS 173 

(120 - 32) (67 - 32) 

Thus A = ^ 19r / V „ 7 ; X .003 X 3375. 

120 — 67 

_ (88X35) x 0Q3 x 3375 _ 

DO 

3080 31185.000 

= ~- X .003 X 3375 = ^ = 588.396 Ans. 

53 od 

Note in the above example there is no sign between 
the brackets. Whenever this is the case (that is, when- 
ever there is no sign between two quantities) the two 
quantities must be multiplied together. 

S 2 
Example. What is the value of jr~ when S = 32 and 

C = 32,X2? 

S*_ 32 X 32 1024 

ihus 8C " 8 X 32 X 2 ~ 512 ~ 2 ' An8 ' 



PART III 

MENSURATION 



TABLE OF DECIMAL EQUIVALENTS 



1 _ 

8 


.125 


A = .01563 


11- = .34375 


ii =.65625 






A - .03125 


|| = .39538 


|| = .67188 






A = .04688 


| =.375 


i| = .6875 






A = .0625 


|| = .39063 


t| = .70313 


1 _ 

4 — 


.25 


A = .07813 


|| = .40625 


If = .71875 






A = .09375 


|| = .42188 


f| = .73438 






e 7 ¥ = .10938 


A - -4375 


| =.75 


3 _ 

8 — 


.375 


§ =-125 


|| = .45313 


|| = .76563 






6 9 4 - = .14063 


i| = .46875 


ff = .78125 






A = -15625 


|| = .48438 


|| = .79688 


1 _ 

2 


.5 


|| = .17188 


i =.5 


H = .8125 






A - .1875 


f| = .51563 


|| = .82813 






|| = .20313 


II = .53125 


|| = .84375 


5 _ 
8 ~ 


.625 


A = .21875 


fl = .54688 


|| = .85938 






If = .23438 


T % = .5625 


i =-875 


3 _ 

4 — 


.75 


| =.25 


f | = .57813 


|| = .89063 






|| = .26563 


3-2 == .oyo/o 


ff = .90625 






j2 r= .^Ol^O 


|| = .60938 


|| = .92188 






|f = .29688 


f =.625 


|| = .9375 


7 _ 
8 


.875 


A = -3125 
|| = .3283 


fi = .64063 


f| = .95313 
fl = .96875 
f| = .98438 
1 =1.00000 



EXPLANATION OF TERMS 



The Point 

A point is that which has position but no dimension. 
An object having a position but no extension. A place 
having a specific position but no size. That is, a point 
has neither length, breadth, nor thickness, hence it has no 
dimension. 

177 



178 A HAXD BOOK FOR MECHANICS 

The Line 

The Straight or Right Line 

Straight means erect, tight, without bend, like a string, 
for instance, which is tightly stretched. A line is length 
without breadth. That is, a line is space of one dimen- 
sion (length). And a line which lies evenly between 
two extreme points is called a straight or right line. 
Xo matter what the position of the points so long as 
they are connected by a line without bend, that line is 
straight. 

Thus, the lines A B, CD, EF are all straight lines. 




Parrallel Lines 

Parallel means alike, similar to, having the same direc- 
tion or course. Hence parallel lines are lines which are 
alike, lines having the same direction or course and 
lying in the same plane as the lines A, B. 



Now if we draw another line, C, having the same 



EXPLANATION OF TERMS 179 

direction and course as the lines A, B, it will be parallel 
to them, thus the three lines A, B, C, are all parallel 
lines. 

, . A 

B 

C 

The Vertical and Perpendicular Line 

A vertical and perpendicular line are one and the 
fame thing. The word vertical relates to the word 
vertea, which means the highest point or summit. A 
vertical line may be thought of as such a line as would 
be formed by a string or line which is suspended from 
some point overhead; such a line would be upright and 
would be perpendicular to the horizon. Thus the line 
A is vertical and is perpendicular to the line C D. 



The Diagonal and Oblique Line 

, Diagonal and oblique have practically the same mean- 
ing. Diagonal means across from angle to angle, oblique 
means aslant, or slanting. By diagonal line, then, is 
meant a slanting line. 
Thus 



180 A HAND BOOK FOR MECHANICS 

The Horizontal Line 

By horizontal line is meant a line that is parallel to 
the horizon; such a line is neither vertical or inclined and 
is represented thus: ' Horizontal. 

The Curved Line 

A curved line is a line whose direction changes con- 
tinually along its path thus 
or 



The Convex Line 

A convex line is a curved, rounded, or arched line, 
which curves away from the line of view. Thus the 
curved part, C, of the figure A, C, B, is said to be convex, 
because it curves away from the ends, A and B, of the 
figure A, C, B. 

c 



A sphere, or circle, looked at from any point without, 
presents a convex surface or line. 



The Concave Line 

The concave line is the reverse of the convex line. 
Concave means incurved. Thus the curved part, E, of 
the figure C, E, K, is said to be concave because it curves 
in. 



EXPLANATION OF TERMS 



181 



A sphere, or circle, when looked at from any point 
within, presents a concave surface or line. 



Examples for Exercise 
What is the name of this 



line? 



Arts. Horizontal line. 
What is the name of that line which lies evenly between 
two points? Ans. Straight line. 

What are these lines called? IZZZZZIZ^IIZ 

Ans. Parallel lines. 
Ans. Vertical or perpen- 
dicular. 

Ans. 
Diagonal or 
oblique. 
Ans. 
Curved line. 



What is the name 
of these lines? 

What is the name 
of these lines? 



What is the name 
of this line? 

What is the difference between a convex and concave 
line? Ans. The convex is outcurved and the concave 
is incurved. 



MENSURATION 

Mensuration is the art of measuring length and volume, 
content, etc. For instance, the art of determining the 
length of a circle or the contents of a cylinder or other 
shaped vessels, and it is necessary that the pupil should 
become thoroughly familiar with, the names of the differ- 
ent shaped surfaces given in this chapter, and at any 
time be able to form a mental conception of same, be- 
cause mensuration treats of the measurement of such 
curved lines, surfaces, and solids. 



182 A HAND BOOK FOR MECHANICS 

The Circle 

The circle is a plane figure formed by a curved line 

called the "circumference," (Fig. 
1), and is such that all right lines 
drawn from a certain point 
within, called the " center," to 
the circumference, are equal to 
each other. 

A radius of a circle is any 

right line drawn from the center 

to the circumference, such as 

FlG - l C D or C A or C B. 

A diameter of a circle is a straight line drawn through 

the center and ending both ways at the circumference, 

as A B. 

To Find the Circumference of a Circle 

Rule. Multiply 3.1416 by the diameter. 

Example. What is the circumference of a circle whose 
diameter is 4 inches. 




Process: 



3.1416 
4 



12.5664 inches. Arts. 

Example. What is the circumference of a circle whose 
diameter is 3J inches? 

L ROCESS '. rt -i a -t r* 

3.1416 

3.5 3+ = 3.5 



157080 
94248 



10.99560 inches. Arts. (Almost 11 
inches 



MENSURATION 183 

Examples for Exercise 

1. What is the circumference of a circle whose diameter 
is 4 inches? Arts. 12.5664. 

2. What is the circumference of a circle whose diameter 
is 4.75? Ans. 14.922600. 

3. What is the circumference of a circle whose diameter 
is 4f ? Ans. 14.922600. 

4. What is the circumference of a circle whose diameter 
is 5.5? Ans. 17.27880. 

5. What is the circumference of a circle whose diameter 
is 6.6? Ans. 20.73456. 

6. What is the circumference of a circle whose diameter 
is 5.6? Ans. 17.59296. 



To Find the Area of a Circle 

By area is meant that which is confined within some 
specific surface space. For in- 
stance, the base, or site, on ^MIIIIIMII^k" ~~f" 
which a building stands. M\ . IK 

Thus, in the following exam- /[ 
pie, which is to find the area 
of a circle whose diameter is 3 v 
inches, the answer tells the \|i| iiijH \\W 

number of square inches con- xH i!i, : jB^ 

tained within a circle whose 
diameter is 3 inches, or the 

number of square inches that would be enclosed on a 
plane surface by a circle whose diameter is 3 inches. 

Example. Find the area of a circle whose diameter 
is 3 inches. (Fig. 2.) 

Rule. Multiply .7854 by the square of the diameter. 
Thus: 



184 A HAND BOOK FOR MECHANICS 

3 .7854 

3 9 

9 square of diameter 7.0686 square inches. Arts. 

Example. The diameter of a circle is 3.5, what is its 
area? 



3.5 


.7854 


3.5 


12.25 


175 


39270 


105 


15708 


12.25 


15708 




7854 



9.621150 square inches. Ans. 

Examples for Exercise 

1. What is the area of a circle whose diameter is 5.5 
inches? Ans. 23.758350. 

2. What is the area of a circle whose diameter is 4 
inches? Ans. 12.5664. 

3. What is the area of a circle whose diameter is 3.75 
inches? Ans. 11.04468750. 

4. What is the area of a circle whose diameter is 2 
inches? , Ans. 3.1416. 

5. What is the area of a circle whose diameter is 6 J 
inches? Ans. 37.0910058750. 

6. What is the area of a circle whose diameter is 4.6 
inches? Ans. 16.619064. 

The Ellipse 

An ellipse may be called a flattened circle, the longest 
diameter of which is called the " transverse axis" (latus 
transversum) and the shortest is called the "conguatt 
axis." (Fig. 3.) 



MENSURATION 



185 




Fig. 3 

To Find the Circumference of an Ellipse 

Rule. Multiply 3.1416 by half the sum of the two 
diameters. 

Example. What is the circumference of an ellipse 
whose diameters are 6 and 4 inches? 

6 

^4 

2)10 

5 X 3.1416 = 15.7080 inches. Arts. 

Example. What is the circumference of an ellipse 
whose diameters are 3| and 7 J? 



3i = 

7i = 



3.75 
7.5 
2 )11.25 
5.625 



3.1416 

5.625 
157080 
62832 
188496 
157080 
17.6715000 inches. Ans. 



To Find the Area of an Ellipse 

Rule. Multiply .7854 by the product of the two 
diameters. 



186 



A HAND BOOK FOR MECHANICS 




Example. What is the area of an ellipse whose 
diameters are 3| and 4j inches? (Fig. 4.) 



3J = 3.5 




.7854 


4± = 4.5 




15.75 


175 




39270 


140 




54978 


Product 15.75 




39270 

7854 

12.370050 square inches. Ans. 


What is the area of 


an ellipse whose diameters are 


5 and 10 inches? 






* 


5 


.7854 




10 


50 




50 


39.2700 square inches. Ans. 



Examples for Exercise 

1. What is the area of an ellipse whose diameters are 
5f inches and 4J inches? Ans. 19.19321250. 

2. What is the area of an ellipse whose diameters are 
7 feet and 9 feet? Ans. 49.4802 square feet. 

3. What is the area of an ellipse whose diameters are 
3.5 feet and 7.5 feet? Ans. 20.616750. 



MENSURATION 



187 



4. What is the area of an ellipse whose diameters are 
3.5 feet and 4.5 feet? Ans. 12.370050. 

5. What is the area of an ellipse whose diameters are 
5 inches and 10 inches? Ans. 39.2700 square inches. 

The Triangle 

A triangle is a figure formed by three right lines joined 
together end to end. The three lines are called its 
" sides." Thus a, b, c, are the sides of the triangle 1. 
(Fig. 5.) 





Fig. 5 

Any three-cornered or three-sided figure body bounding 
a three-sided space is called a triangle. 

A triangle whose three sides are unequal is called a 
Scalene. (Triangle 1.) 

A triangle having two of its sides equal is called an 
" Isosceles Triangle." (Triangle 2.) 

And a triangle all of whose sides are equal is called an 
" Equilateral Triangle.'- (Triangle 3.) 



To Find the Area of a Triangle 

Rule. Multiply the base by half the perpendicular 
height. 

Example. What is the area of a triangle whose base 
is 4 feet and whose height is 4 feet? 



188 A HAND BOOK FOR MECHANICS 

Half the height = 2 feet; thus 2x4 = 8 square feet. 

Example. Find the area of a triangle whose base is 
12 inches and whose height is 8 inches. (Fig. 6.) 




Fig. 6 



Half height = 4 inches; then 4 X 12 = 48 square inches 

Example. Find the area of a triangle whose base is 
12 inches and whose height is 4 inches. (Fig. 7.) 




Fig. 7 



Half the height = 2 inches; then 2 X 12 = 24 square 
inches. 

Example. Find the area of a triangle whose base is 
15 feet and whose height is 9 feet. (Fig. 8.) 

Half the height = 4.5; then 4.5 X 15 feet. 



MENSURATION 



189 



15 
^5 
75 
60 
67.5 square feet. 




Fig. 8 



Examples for Exercise 

Find the area of triangles having the dimensions below. 

1. Base 15 feet, height 4.5. Arts. 33.75 square feet. 

2. Base 14.75 feet, height 61 feet. Arts. 46.09375 
square feet. 

3. Base 8 inches, height 4 inches. Ans. 16 inches. 

4. Base 27.5 feet, height 5 feet. Ans. 68.75 square feet. 

The Square 



A square is an area 
bounded by four equal 
sides. (Fig. 9.) A figure all 
of whose sides are equal 
and all of whose angles are 
right angles. 




Fig. 9 



190 A HAND BOOK FOR MECHANICS 

To Find the Area of a Square 

Rule. Multiply the base by the height, or the length 
by the breadth. 

Example. What is the area of a square whose base is 
3 inches? (Fig. 9.) 

3 

3 

9 square inches. Ans. 

Example. What is the area of a square whose base 
is 3.5 feet? 

3.5 

3.5 



175 
105 
12.25 square feet. 



Examples for Exercise 

Find the area of the following squares: 

1. Of a square whose base is 9 feet. Ans. 81 square 
feet. 

2. Of a square whose base is 36 inches. Ans. 9 square 
feet. 

3. Of a square whose base is 3 feet. Ans. 9 square feet. 

4. What is the area of a square whose side is 2.5 feet? 
Ans. 6.25 square feet. 

The Oblong 

An oblong is a figure whose sides are perpendicular to 
its base and whose length is greater than its breadth. 
(Fig. 10.) 



MENSURATION 



191 




Fig. 10 

To Find the Area of an Oblong 

Rule. Multiply the length by the breadth. 

Example. What is the area of an oblong whose length 
is 7 feet and whose height is 4 feet? (Fig. 10.) 

7 = base 
4 = height 
28 square feet. Ans. 

Example. What is the area of an oblong whose base 
is 12 feet and height 4 feet? 



12 
_4 

48 square feet. Ans. 



Examples for Exercise 

Find the area of the following oblongs: 

1. Base 12 feet, height 6 feet. Ans. 72 square feet. 

2. Base 5.5 feet, height 3.5 feet. Ans. 19.25 square 
feet. 

3. Base 9f inches, height 5 inches. Ans. 48.75 square 
inches. 

4. Base 15 feet, height 2.5 feet. Ans. 37.5 square 
feet. 



192 



A HAND BOOK FOR MECHANICS 

The Rhomboid 



A rhomboid is a four-sided figure whose opposite sides 
are parallel, whose dimensions are greater one way than 
another, but whose ends are not perpendicular to its base. 

To Find the Area of a Rhomboid 

Rule. Multiply the base by the perpendicular height. 

Example. Find the area of a rhomboid whose base is 
9 feet and perpendicular height 3 feet. (Fig. 11.) 




Fig. 11 



9 
_3 
27 square feet. Arts. 

Example. Find the area of a rhomboid whose base is 
6 feet and whose height is 3^ feet. 

6. 

3J5 

30 
18 
21.0 square feet. 

What is the area of a rhomboid whose base is 3 feet 
and whose perpendicular height is 4 feet? (Fig. 12.) 



MENSURATION 



193 




Fig. 12 

3 

_4 

12 square feet. Arts. 

Example. Find the area of a rhomboid whose base is 
2 feet and whose height is 5^ feet. (Fig. 13.) 




*—i 



Fig. 13 
2. 

10 
10 



11.0 square feet. Arts. 



194 



A HAND BOOK FOR MECHANICS 

The Trapezoid 



A trapezoid is a plain four-sided figure, having two of 
its opposite sides parallel and the other two not so. 
(Figs. 14 and 15.) 




Fig. 14 



Fig. 15 



To Find the Area of a Trapezoid 

Rule. Multiply half the sum of the two parallel sides 
by the distance between them. 

Example. What is the area of the following trapezoid? 
(Fig. 16.) 

U = I" J 




Fig. 16 



Here, the lines 8" and 6" being the parallel sides, we 
multiply half their sum by the perpendicular distance 
between them, which is 3 inches. Thus: 



8 

_6 

2)14 



7 X 3 = 21 square inches. 



MENSURATION 



195 



Example. What is the area of a trapezoid whose 
parallel sides are 5.5 feet and 9.75 feet, and the distance 
between them is 2 feet? (Fig. 17.) 




Fig. 17 



Thus; 



9.75 
5.5 
2)15.25 



7.625 X 2 = 15.250 square feet 



Examples for Exercise 

1. What is the area of a trapezoid, the longer of the 
two parallel sides being 130 feet, the shorter 100 feet, 
and the height 80 feet? Ans. 9200 square feet. 

2. What is the area of a trapezoid, the longer of the 
two parallel sides being 25 feet, and the shorter 23 feet, 
and the width being \\ feet? Ans. 30 square feet- 



The Trapezium 

A trapezium is any plane figure contained by four 
straight lines, no two of which are similar. (Figs. 18, 19, 
20.) 



196 



A HAXD BOOK FOR MECHAMCS 




Figs. 18. 19. 20 



To Fixd the Area of a Trapezium 



Rule. Divide the trapezium into two triangles, by 
connecting its opposite angles, making this line of division 
the base line of both triangles. Measure the perpendicu- 
lar height of each triangle from this base line. Then 
find the area of each triangle, add both together, and the 
sum is the area of the whole figure. 

Example. Find the area of the following trapezium. 
(Fig. 21.) 




Fig. 21 



MENSURATION 



197 



Thus 2)4 

2 X 12 =24 sq. ft., area of triangle C 
2)3 

1.5 X 12 = 18.0 sq. ft., area of triangle B 
42.0 sq. ft. Ans. 
Here we first of all divide the figure in two by the 
dotted line (A); we then have two triangles, B and C, 
and this line (A) proves the base for both. We then 
find the area of each triangle separately, add them both 
together, and the sum 42.0 equals the area of the whole 
figure. 

Example. What is the area of the following trape- 
zium? (Fig. 22.) 

Thus 2)8 

4X20= 80 
2)4 
2 x 20 = jtO 

120 sq. ft. Ans. 




To Find the Surface of a Cylinder 

Rule. Multiply the diameter by 3.1416 and multiply 
the product by the height. 



198 



A HAND BOOK FOR MECHANICS 



Example. Find the number of square inches in sur- 
face of a cylinder of the following dimensions. (Fig. 23.) 




Fig. 23 



Thus 4" = diameter 
And 10" = height 

Then 4" X 3.1416 = 12.5664 circumference 

10 

125.6640 sq. in. Arts. 

Example. What is the surface of a cylinder whose 
diameter is 9 inches and height 15 inches? 

9 X 3.1416 = 28.2744 =• circumference 

Thus, 28.2744 X 15 = 424.1160, area of surface in 
square inches. 



MENSURATION 



199 




The Sphere 

A sphere is a body bounded by a curved surface, all 
parts of which are an equal 
distance from a certain point 
within, called the " center." 
The diameter of a sphere is a 
straight line drawn through 
its center, terminating both 
ways in the surface. 

The circumference of a 

sphere is the greatest distance 

around the sphere. (Fig. 24.) 

F v & ' Fig. 24 

To Find the Surface of a Sphere 

Rule. Multiply 3.1416 by the square of the diameter. 

Example. What is the surface of a sphere whose 
diameter is 6 feet? 

Thus 6 

_6 

36 = sq. of diameter X 3.1416 = Answer. 
3.1416 

36 

188496 
94248 



113.0976 area of surface in square feet. Arts. 
Example. What number of square inches of gold leaf 
will cover a sphere whose diameter is 18 inches? 



18 
18 



144 
18_ 

324 square of diameter 



3.1416 
.324 

125664 
62832 
94248 
1017.8784 sq. in. Arts. 



200 A HAND BOOK FOR MECHANICS 

The contents of a sphere are equal to the product of 
| of 3.1416 multiplied by cube of diameter. 

Example. How many cubic feet of gas will fill a 
spherical balloon whose diameter is 6 feet ? 

J of 3.1416 X6 3 = 113.097 cubic feet. 4ns. 

VOLUME MEASURE AND CONTENTS OF SOLIDS 

Rectangular Solids 
To Find the Contents of a Rectangular Solid 

A rectangular solid is a figure bounded by six rect- 
angles. (Fig. 25.) 
The dimensions of a rectangular figure are its length, 

breadth, and thickness, and the 
contents or volume of a rectan- 
gular figure is the space contained 
within its bounding surfaces. 

The volume or content of any 
rectangular is determined by the 
following rule: 

Rule. Multiply the length, 

breadth, and height together. 
Fig. 25 

Example. What is the con- 
tent of a rectangular solid whose length is 3 feet, breadth 
2 feet, height 2 feet? 

3 
2 

6 
_2 

12 cubic feet. Ans. 

Example. How many cubic feet in a rectangular 
figure, each of whose sides measures 3 feet? 




VOLUME MEASURE AND CONTENTS OF SOLIDS 201 

3 

3 

9 

_3 

27 cubic feet. Ans. 

Example. How many cubic feet of water will a vessel 
hold whose inside dimensions are, length 5 feet, breadth 
4 feet, and depth 3 feet? 

5 X 4 X 3 = 60 cubic feet of water 



The Cylinder 

A cylinder is a round body of a uniform diameter, 
whose bases are equal and parallel 
circles. (Fig. 26.) 

To Find the Cubic Contents of a 
Solid Cylinder 

Rule. Find the area of the base 
and multiply this by the height or 
length. 

Example. What are the cubic con- 
tents of a cylinder whose diameter is 
2 feet and whose height or length is 7 feet? 

Thus 2 
2 

4 X .7854 is .7854 
4 




Fig. 26 



3.1416 = area of base in sq. ft. 
7 = height of cylinder 
21.9912 cubic feet. Ans. 



202 



A HAND BOOK FOR MECHANICS 



Example. How many cubic feet of water will a 
cylindrical shaped vessel hold, whose inside dimensions 
are, diameter 4 feet, and height 1\ feet? 

4 .7854 

_4 16 

16 47124 

7854 
12.5664 = number of sq. ft. in area of base 

7.5 = height 
628320 
879648 



94.24800 cubic feet. Arts. 



The Pyramid 

A pyramid is a body whose base is a polygon but whose 

sides are all triangles, meeting 
at one point (A), called the 
vertex of the pyramid. (Fig. 
27.) 

Polygon means manycurves 
or many angles. 

A polygon, then, is a plane 
figure bounded by three, four, 
five, six, or any number of 
sides. 

Accordingly, the base of a 
pyramid may be triangular, 
square, pentagonal, etc., and 
pyramids are named for the 
figure or shape of their base. 

A pyramid having a three- 
curved or triangular base is called a trianglor pyramid, 
and one having a square base is called a square pyramid, 




Fig. 27 



VOLUME MEASURE AND CONTENTS OF SOLIDS 203 

and one having a five-sided or pentagonal base is called 
a pentagonal pyramid. 

To Find the Cubic Contents of a Pyramid 

Rule. Multiply the area of the base by one third of 
its perpendicular height (altitude). 

Example. How many cubic inches are there in a pyra- 
mid of the following dimensions: base 3 inches square, 
and height (altitude), 12 inches? 

3 
3 

9 = square inches area of base 



3)12 

4 = one third of altitude 

9 
_4 

36 cubic inches. Arts. 

Example. How many cubic inches are there in a 
pyramid whose base is a triangle, each of whose sides 
measures 3 inches and whose height is 15 inches? Thus: 

3 

15 = one half of perpendicular height 
4.5 = area of triangle base 
5 = one third of height of pyramid 



22.5 = number of cubic inches. Ans. 

Process. We first find the area of the triangular base, 
according to the rule given for finding the area of a 
triangle, which we multiply by one third of the per- 
pendicular height of the pyramid, the product 22.5 thus 
obtained being the answer. 



204 



A HAND BOOK FOR MECHANICS 



The Cone 

A cone is a body whose base is a circle and whose 
convex surface tapers uniformly to 
a point, "A," called the vertex of 
the cone. (Fig. 28.) 

To Find the Cubic Contents 
of a Cone 

Rule. Multiply the area of the 
base by one third the perpendicular 
height. 

Example. Find the cubic inches 
in a cone whose base is 3 inches 




Fig. 28 



diameter and whose height is 15 inches. 



.7854 

9 

7.0686 

_5 

35.3430 



area in square inches of base. 

one third of perpendicular height of cone. 

cubic inches contained in cone. 



Example. How many cubic inches of water will a 
cone-shaped vessel hold, the diameter of whose base is 
3^ inches and whose height is 12 inches? 



3.5 


.7854 


3.5 


12.25 


175 " 


39270 


105 


15708 


12.25 


15708 




7854 




9.621150 


» 


4 



area of base. 

one third of perpendicular height. 
38.484600 cubic inches of water. Arts. 



VOLUME MEASURE AND CONTENTS OF SOLIDS 205 

The Frustum 

A frustum is that part of any solid figure which is 
between two planes, either parallel or inclined to each 
other. It is the part of a solid which remains after 
cutting off the top part by a plane parallel to the base. 

The "frustum" actually means, a piece; particularly, 
a remaining piece of something of which a part is lacking. 

A frustum of a cone, or pyramid, then, is the part 
which remains after cutting off the top by a plane parallel 
to the base. (Figs. 29 and 30.) 





Fig. 29 



Fig. 30 



To Find the Cubic Contents of a Frustum of a Cone 

Rule. Find the sum of the squares of the two diam- 
eters, add on to this the product of the two diameters, 
then multiply by .7854 and then multiply by one third 
the height. 

Example. Find the cubic contents of a frustum of a 
cone whose large diameter is 12 inches, small diameter 
6 inches, and height 4 inches. 



206 



A HAND BOOK FOR MECHANICS 



12 
12 



144 = square of large diameter 36 = sq. of small 
36 [diameter. 

180 = sum of squares of both diameters 
72 = product of both diameters 
252 
.7854 12 

6 
72 = product of both diameters. 



3)4 



1.33 = one-third of thickness. 



1008 
1260 
2016 
1764 

197.9208 
1.33 
5937624 
5937624 
1979208 
263.234664 cubic inches. Ans. 

Example. How many cubic inches in the body shown 
in Figure 31? 




Fig. 31 



VOLUME MEASURE AND CONTENTS OF SOLIDS 207 

20 
JO 

400 = square of large diameter. 
100 



500 = 


sum of square of both 


200 = 


product of both diamel 


700 




.7854 


2800 




3500 




5600 




4900 


• 


549.7800 




2 = 


= one-third of thickness. 


1099.5600 cubic inches. Ans. 


10 




10 




100 = 


= sq. of small diameter. 


20 




10 





200 = product of both diameters. 
3)6 

2 = one-third of thickness. 

To Find the Cubic Contents of a Frustum of a 

Pyramid 

Rule. The sum of the areas of the two bases added 
to the square root of their product, multiplied by one 
third of the altitude. 

Example. What are the contents of the frustum of a 
square pyramid whose height is 30 feet, and whose side 
at the base is 20 feet and at the top ten feet? (Fig. 32.) 



208 



A HAND BOOK FOR MECHANICS 




Fig. 32 



Process: 

20 X 20 = 400; 10 X 10 = 100; 400 X 100 
V40000 = 200; 200 + 400 + 100 = 700; 
30 -T- 3 = 10; 700 X 10 = 7000 cubic feet. 



= 40000; 



PART IV 

WEIGHT, SPECIFIC GRAVITY, HOW THE DIMEN- 
SIONS, MEASUREMENTS AND WEIGHT OF 
DIFFERENT SHAPED VESSELS ARE FOUND, 
AND HOW 7 THE WEIGHT OF DIFFERENT 
PARTS IS FOUND 



WEIGHT 

By weight, is meant the heaviness of a body, the 
downward force a body hasj which force is created by 
the action of a force called the " force of gravity," working 
upon all its particles. 

The " force of gravity" is that attraction which the 
earth exerts upon all bodies near it, tending to draw them 
toward its center. When this force is resisted it gives 
rise to pressure, which is called weight. Weight, there- 
fore, is due to and is the effect of resisted gravity. 

SPECIFIC GRAVITY 

In nature all substances have, under the same condi- 
tions, a weight " specific or peculiar" to themselves. 
This is due to the fact that like volumes of different 
materials contain variable amounts of matter. 

There is, for instance, more matter contained in a 
cubic inch of lead than in a like volume of wood, and the 
greater the density of the body or, in other words, the 
greater the amount of matter contained within any 
specific space, the greater the weight, for the reason that 
the action of gravity on all bodies is proportional to 
their volume or density. 

Therefore, under the same conditions, a cubic inch of 
lead will weigh more than a cubic inch of wood, and the 
comparative weights of equal volumes of different sub- 
stances are called their "specific" gravities; and the 
standard of reference is pure water at a temperature of 
60° Fahrenheit. 

In other words, the " specific gravity" of a body is its 

211 



212 A HAND BOOK FOR MECHANICS 

relative weight, that is, the number of times it is " heavier" 
or "lighter" than a body of the same size of a different 
substance, and in order to determine the relative weights 
of equal sized bodies of different composition a standard 
is taken, which is pure water at the temperature of 
60° Fahrenheit. 

There are different methods used for taking the specific 
gravities of bodies, one of which, called the "Hydrostatic 
Balance/' is here described. By this method a regular 
balance is used, having two scale pans. The body whose 
specific gravity is to be taken is placed on one of the 
scale pans, and on the other pan sufficient known weight 
is placed to exactly counterbalance it. This will give 
the weight of the body in the air. 

The body is then taken and suspended by a thin wire 
from the bottom of the scale pan and completely sub- 
merged in pure water. When thus submerged the body 
is counterbalanced by placing weights in the other scale 
pan and, because of the buoyancy of the water working 
on the body thus submerged, it will be seen to weigh 
less in water than in air. 

Now, in taking the specific gravity of solids, advantage 
is taken of the important fact that when a solid is wholly 
submerged in water it displaces a volume of that liquid 
exactly equal to its own volume, and the solid appears 
to lose its weight, that is, it is supported by the sur- 
rounding water with a force exactly equal to the weight 
of the water displaced; hence a body will weigh less in 
water than in air, and the difference of its weight in 
water from that of its weight in air must be the weight 
of an equal volume of water. 

For instance, if a piece of glass is found to weigh, in 
air, 577 grains, and when suspended by a fine wire from 
the bottom of the scale pan and immersed in a vessel of 



SPECIFIC GRAVITY 213 

pure water it is found to weigh 399.4, the difference 
between the weight in air (577) and the weight in water 
(399.4) is the weight of the volume of water displaced 
by the glass. Therefore: 

577.0 

399.4 

177.6 

is the difference, which difference is the weight of the 
water displaced. 

Hence the rule for finding the specific gravities of solids: 
Weigh the solid in air and then in pure water (distilled 
water) and divide the weight of the body in air by the 
difference between the weights in air and water. 

Hence, if a piece of glass is found to weigh 1154 grains 
in air and 798.8 grains in water, the specific gravity of 
the glass will be determined by dividing 1154.0 by the 
difference between 1154.0 and 798.8. Thus: 



Ans. 



1154.0 




7S8.8 




355.2 


355.2)1154.0(3.248 




10656 




8840 




7104 




17360 




14208 




31520 




28416 



This shows us, then, that glass is 3.248, etc., times 
heavier than water. That is, a (cubic inch), for instance, 
of glass would weigh over three times as much as a cubic 
inch of water. 

Example. If a piece of marble weighs 48.0 grains in 
air and 31.0 grains in water, what is its specific gravity? 



214 A HAND BOOK FOR MECHANICS 

48.0 
31.0 

17.0 = difference between weight in air and water 
17.0)48.0(2.823, etc. Arts. 
340 

1400 
1360 

400 
340 

600 
510 

90 

The specific gravities of liquids are ascertained by an 
instrument called the " Hydrometer." 

A Table of Specific Gravities of Liquids and Solids 

NAME SPECIFIC GRAVITY 

Air 0.001228 

Pure Water 1.0000 

Sea Water , 1.029 

Alcohol ." .79 

Linseed Oil 0.9347 

Olive Oil 0.9176 

Solids 

Gold 19.3 

Lead 11.4 

Copper 8.767 

Brass 8.384 

Cast Iron (Average) 7 . 1 10 

Wrought Iron (Average) 7 . 690 

Steel 7.780 

Tin 7.293 

Zinc • 7.215 



SPECIFIC GRAVITY 215 

A cubic foot of pure water weighs 62.5 pounds. There- 
fore, to find the weight of anything contained in the above 
table, multiply 62.5 (the weight of a cubic foot of pure 
water) by the specific gravity of the given body. 



Example. 


What 


is 


the weight of a cubic foot of 


alcohol? 






.79 
62.5 
395 
158 
474 

49.375 lbs. Arts. 



Example. What is the weight of 3 cubic feet of 
alcohol? 

.79 
62.5 

395 
158 
474 



49.375 = weight of 1 cubic foot 
3 



148.125 = weight of 3 cubic feet 

Example. What is the weight of a cubic foot of cast 
iron? 

7.110 
62.5 
35550 
14220 
42660 



444.3750 lb. Ans. 
Example. What will 2 cubic feet of steel weigh? 



216 A HAND BOOK FOR MECHANICS 

7.780 
62.5 



38900 
15560 
46680 

486.2500 lb. = weight of 1 cubic foot. 
2 

972.5000 lbs. Ans. 

Example. If one cubic foot of steel weighs 486 pounds, 
how many cubic feet are there in a ton (2000 pounds)? 

486)2000 (4.115, etc., cubic feet of steel 
1944 in a ton 





560 






486 






740 


•< 




486 






2540 






2430 


• 




110 




Example. 


How much does a 


cubic foot of wrought 


iron weigh? 


62.5 
7.690 

56250 
3750 
4375 





480.6250 lb. Ans. 

In practice 480 pounds is called the weight of a cubic 
foot of wrought iron. 

Example. How many cubic feet of wrought iron are 
there in a ton? 



SPECIFIC GRAVITY 

480)2000(4.16 
1920 

800 
480 



217 



3200 

2880 



3200 
4.16 cubic feet = 1 ton. 



Ans. 



Example. What is the weight of a cubic inch of 
wrought iron? 

1 cubic foot = 1728 cubic inches 
480 pounds = weight of 1 cubic foot 

1728)480.0(.27 of a pound. Ans. 
3456 

13440 
12096 



1344 

Example. One cubic inch of brass weighs what ? 

1728)521.75(.313. Ans. 
5184 



3350 
1728 



6220 



Example. How many cubic inches of wrought iron 
weigh 1 pound? 

480)1728(3.6 cubic inches = 1 lb. 
1440 



2880 
2880 



Example. How many cubic inches of fresh water 
weigh 1 pound? 

One cubic foot of fresh water weighs 62.5 pounds. 



218 A HAND BOOK FOR MECHANICS 

62.5)1728.0(27.656 
1250 
4780 
4370 



4100 
3750 



3500 
3125 

3750 

3750 

0000 

27.656 cubic inches of fresh water = 1 pound. Ans. 
Example. How much will 2 cubic feet of lead weigh? 

62.5 
11.4 



2500 
625 
625 



712.50 = weight of 1 cubic foot 

'2 



1425.00 = weight of 2 cubic feet 

From the above examples we learn how the weight of 
anything may be found when its volume and specific 
gravity are known. Further on examples will be given 
showing how to find the weight of things whose specific 
gravity is known, but whose dimensions are given only, 
by which its volume may be determined, but whose 
volume is not given. 

These examples are given so as to instruct the pupil in 
figuring the sizes of different parts of machinery, etc., 
and also to show how to figure the weights of the different 
parts. 



CIRCULAR MEASURE 219 

CIRCULAR MEASURE 

The diameter of a circle is equal to the circumference 
divided by 3.1416. 

Example. What is the diameter of a circle whose 
circumference measures 12f inches? 

12} = 12.75 
Then 12.75 -f- 3.1416 is 

3.1416)12.7500(4.058 
125664 

183600 
157080 



265200 
251328 



The diameter, then, of a circle whose circumference is 
3J = 4.058 inches. 

Example. A piece of shafting measures 14^ inches 
around its circumference, — what is its diameter? Ans. 
4.535, say 4^ inches. 

Example. What is the diameter of a shaft whose 
circumference measures 14J inches? Ans. 4.535. 

Example. What is the diameter of a shaft whose circum- 
ference measures 12.5664 inches around? Ans. 4 inches. 

Example. A tube is 
made of two plates hav- 
ing flanges A, B, as shown 
in Fig. 33. The outside 
diameter of tube is 3 feet 
and each flange measures 
4 inches; what is the full 
length of each plate be- 
fore being bent? 




Fig. 33 



220 A HAND BOOK FOR MECHANICS 

Reduce first the diameter (3 feet) to inches. 

Thus 3 X 12 = 36 inches 
Then 36 

3.1416 
216 
36 
144 
36 
108 



113.0976 



equals circumference in inches, to which add the 4 flanges, 
each of which is 4 inches long, equals 16 inches, then 
divide by 2, because there are two plates, and the 
quotient is the length of each plate. Thus: 

113.0976 
16 



2 )129.0976 

64.5488 inches = length of each plate 
About 5.379 feet. Ans. 

Example. The outside diameter of a tube which is 
made of three plates is 4 feet, each plate has a lap of 
1 J inches, what is the full width of each plate? (Fig. 34.) 

If = 1.75 

4 X 12 = 48 = diameter in inches 

48. 
3.1416 



288 
48 
192 
48 
144 



150.7968 = circumference in inches 



CIRCULAR MEASURE 



221 




Fig. 34 

Now, as this tube is made of three equal sections, the 
whole circumference divided by 3 will equal the length 
of each section from A to B, and to this length must be 
added the length of the laps, If inches, and the sum 
will equal the whole width of each plate. 



Thus: 



150.7968 



+ 1.75 = 52.0156 inches. Arts. 



Example. What is 
the full length in inches 
of a brass band, \ inch 
thick, which goes around 
a funnel whose diameter 
is 5i feet? The end of 
the band laps 3 inches. 
(Fig. 35.) 




Fig. 35 



222 



A HAND BOOK FOR MECHANICS 



5J = 5.25 
5.25 X 12 X 3.1416 = 197.920 circumference in inches 

of strap. 
Then 197.920 + 3 = 200.920 inches, or 16.743 feet, 

Example. If a circular plate, 5 feet in diameter, has 
to have some holes bored in it, 2\ inches from the edge 
of the plate to the center of the holes, what will be the 
diameter of the circle on which holes are bored? (Fig. 36.) 




Fig. 36 

The circle on which the holes are bored is 2\ inches 
inside the outer diameter, of plate, and therefore five 
inches less in diameter. Diameter of plates = 5 feet = 
60 inches 

Then 60 — 5 = 55 inches. Arts. 




l'IG. 61 



Example. What is the 
full length around and 
across the base of an 
arched piece of metal like 
Figure 37. The arch 
equals \ circle. 

Divide circumference of 



CIRCULAR MEASURE 



223 



circle, whose radius is 3 inches (diameter 6 inches) by 2, 
to which add 6 inches (distance across base). 



Thus 



3.1416 X 6 



+ 6 = 15.4248 inches. Ans. 



Example. What is the outside circumference of a 
ring whose outside diameter is 4 feet? 

And what is the inside circumference when the diam- 
eter is 2 J feet, and what width of ring? (Fig. 38.) 




Fig. 38 



3.1416 X 4 = 12.5664 feet, outside circumference 
3.1416 X 2.5 = 7.85400 feet, inside circumference 
4 — 2.5 = 1.5 -T- 2 =.75 feet, width of ring. Ans. 

Example. The circumference of 
a flange is 60 inches and in it there 
are a number of holes 4 inches from 
the outside of flange to center of 
holes. What is the diameter of cir- 
cle on which holes are drilled? (Fig. 
39.) 

The first thing we do is to find the p IG . 39 

diameter of the flange, from which we subtract twice the 
distance from the outside of flange to center of holes. 




224 



A HAND BOOK FOR MECHANICS 



Thus 



60 



3.1416 



4X2= Ans. 



3.1416)60.0000(19.095 = diameter of flange 
31.416 

285840 

282744 



309600 

282744 



16856 

19.095 - 8 = 11.095. Ans. 

Then 11.095 is the diameter of circle on which holes 
are drilled. 

Square Measurements 

Note: one dash after a dimension represents feet 

Thus 23' = 23 feet, and 6' = 6 feet 
And two dashes after a dimension represents inches. 
Thus 23" = 23 inches, and 6" = 6 inches 




Fig. 40 

Example. Find the surface in square feet, of a sheet 
of metal of the following dimensions. (Fig. 40.) 

This may be done by adding the ends A and B to- 
gether and dividing by 2, which will give the mean 
width, which multiply by the length, 3', 6". 



Thus 



CIRCULAR MEASURE 
2' 4" + 4" 32" 



225 



16 mean width 



Thus 16 X 3' 6" = 16 X 42" = 672 square inches, or 
4.6 square feet, or it may be done by adding together 
the area of the triangle, A, E, C, and the area of the 
parallelogram, A, E, B, K. 

rpi *!.'*■ 1 2 ' X ^ 6 " 24 " X 42 " KiXA 

Ihus the triangle = , or = 504 

square inches. 

And the parallelogram = 4" X 42" = 168 square inches 
144 square inches == 1 square foot 

• Then 504 
And 168 

672 sq. in., or 4.6 sq. ft. Ans. 

144)672(46. 
5/o 

"960 
864 

96 

Example. Find the measurement in square feet, of 
a sheet of metal of the following dimensions. (Fig. 41.) 
Answer: 10^ square feet. 




Fig. 41 



226 A HAND BOOK FOR MECHANICS 

^-2 = 10i Am. 

Example. What is the area of a sheet of metal of 
the following shape and dimensions. (Fig. 42.) Answer: 
88.3575 square inches. 




Fig. 42 



15" X 15" X .7854 



= 88.3575 square inches. Ans. 



The area of the whole circle would equal 176.7150 
square inches, therefore the plate being \ circle equals 
88.3575 square inches. 




Example. What is the area cf a sheet of metal of 
the above shape and dimensions. (Fig. 43.) 
The circular end of plate equals \ circle. 



CIRCULAR MEASURE 



227 



Thus 



5 

25 X .7854 



= 9.8125 sq. ft. 



Find now the area of the remainder of the sheet, 
which add to (9.8125) the area of the circular part. 

Thus 9.8125 + 70 = 79.8125 sq. ft. Ans. 

Example. Find the surface in square feet of a piece 
of plate of the following dimensions. (Fig. 44.) 



-36 





Fig. 44 

31.8087 *= Area of curved part. 
288 = oblong part. 
20.25 = triangular part. 

Thus 31.8087 + 288 + 20.25 = 340.058 square inches., 
or 2.361 square feet. Ans. 

Example. What is the number of square inches in 
the outside surface of a cylinder whose outside diameter 
is 5 inches and height 15 inches. (Fig. 45.) 



228 



A HAND BOOK FOR MECHANICS 




-5— 




Fig. 4o 

3.1416 
5 

15.7080 = circumference 
15 = height 

785400 
157080 



235.6200 sq. in. Ans. 
Example. What is the number of square inches in 
the outside surface of an elliptical tube whose diameters 
are 2 and 4 inches and whose length is 12 inches? (Fig. 46.) 




Fig. 46 



CIRCULAR MEASURE 



229 



3.1416 
3 



half sum of diameters 



9.4248 = circumference 
12 

113.0976 sq. in. Ans. 

Circular Areas 

Example. Find the area of metal in a ring whose outer 
diameter is 13 inches 
and inner diameter 9 
inches. (Fig. 47.) 

First find the area 
of the outer diameter. 
Next find the area of 
the inner diameter. 
Then their difference is 
the required area of 
metal. 




± 



Fig. -.7 



Thus 13 X 13 X .7854 = 132.7326, area of large diameter, 
and 9 X 9 X .7854 = 63.6174 , area of small diameter. 
Difference, 69.1152, area of metal. 

Example. What area of metal is there in a section 

3 inches 
in diameter? (Fig. 48.) 




j— of a round bar of iron 

f 



3 

3 

9 

.7854 X 9 = 70.686 area 



_t_. 



Fig. 48 



Example. What area of metal 
is there in a section of a tube 
whose inside diameter is 5 inches and the thickness of 
tube is \ inch? (Fig. 49.) 



230 



A HAND BOOK FOR MECHANICS 



In this case, as the inner diameter is 5 inches and the 
tube is J inch thick, the outside diameter must be 5" + 
i + 1 = 5.5. 



Jg THICK 




Fig. 49 

Now find the difference between the two areas. 
5 

25 X .7854 = 19.6350 area of inside diameter. 

5.5 
5.5 
275 

275 

30.25 X .7854 = 23.758350 area of outside diameter. 
19.6350 



4.123350 area. Ans. 



Example. The inner diameters of an elliptical tube 
are 3 feet and 4 feet, and the thickness of metal is J inch, 
what area of metal is there in a section of the tube? 
(Fig. 50.) 

As the inner diameters are 3 feet and 4 feet, that is, 
36 inches and 48 inches, and the metal is J inch thick, 
the outside diameters must be 



CIRCULAR MEASURE 



231 



Conguatt axis (outside) \ + 36 + \ = 36^ 
Transverse axis (outside) } + 48 + \ = 48^ 

Inside diameters 3 feet 
And 4 feet 



36.5 inches 

48.5 inches 
36 inches 
48 inches 



Find the difference between the two areas and the 
answer is the area of metal in a section of the tube. 




Fig. 50 

36.5 X 48.5 X .7854 

36 X 48 X .7854 

Arts. 



1390.354 
1356.971 




Fig. 51 

Example. If the outer diameter of a metal ring is 
21 inches and the inner diameter is 19 inches, and the 
ring has 4 quarter-inch holes in it, what is the area ot 
the metal? (Fig. 51.) 



232 A HAND BOOK FOR MECHANICS 

First find the area enclosed by the outer diameter, 
which is 346.3614 square inches. 

Next find the area enclosed by the inner diameter 
which is 283.5294 square inches. Then find their differ- 
ence. Thus: 

346.3614 

283.5294 

62.8320 

From which subtract the sum of the areas of the four 
^-inch holes and the difference is the required area, \ = .25. 



.25 


62.8320 


.25 


.1965000 


125 


62.6355 = are 


50 




.0625 




.7854 




2500 




3125 




5000 




4375 




.04908750 = 


area of \" hole 


4 





. 19635000 = area of 4 holes 



To Find the Diameter of a Circle when the 
Area is Given 

Rule. Divide area by .7854, then find the square of 
the quotient obtained, which will equal the diameter. 

Example. What is the diameter of a circle whose 
area is 190.0668 square inches? 



CIRCULAR MEASURE 233 



.7854)190.0668(242 






15708 






32986 






31416 






15708 






15708 


2,42(15.5 
1 

25)142 
125 

305)1700 
1525 






17500 






15.5 etc. inches. 


Ans 



Example. If the area of a shaft is 3.1416 square 
inches, what is its diameter? 

.7854)31416(4 4(2 inches. Ans. 
31416 4 

Example. If the area of a shaft is 6.2832 square 
inches, what is its diameter? 

.7854)62832(8 8(2.828 etc. 

62832 4 

48)400 
384 



562)1600 
1124 



5648)47600 
45184 

Example. If the area of a circular plate is 706.86 
square inches, what is its diameter? Ans. 30 inches. 

Note. The diameter of a circle equals the square root 
of the area multiplied by 1.12838. However, the diam- 



234 



A HAND BOOK FOR MECHANICS 



eter is determined usually when the area is given by 
the process worked out in the example. 



Measurements and Weights of Tanks. Examples 
Showtng how to find the Cubic Capacity and 
Weight of Rectilinear and Circular Vessels 
and Tanks 

Note. The U. S. Standard gallon measures 231 cubic 
inches and contains 8| pounds of distilled water (pure 
water). A cubic foot of water (fresh) weighs 62 \ pounds 
and contains 1728 cubic inches, or nearly 1\ gallons 
U. S. standard. 




Fig. 52 

Rule. To find the capacity of four-sided vessels in 
gallons, find the cubical contents by multiplying the 
length, breadth, and height in inches, and divide the 
product by 231. 

Example. If the inside dimensions of a tank are 
4 feet wide, 3 feet deep, and 12 feet long, how many 
gallons of water will it hold? (Fig. 52.) 

3 feet = 36 inches. 

4 feet = 48 inches. 
12 feet = 144 inches. 



CIRCULAR MEASURE 235 

Then 36 X 48 X 144 = 248832 cubic inches. 

231)248832(1077, etc., gallons. 
231 

1783 
1617 

1662 

1617 

Example. How many gallons will a tank of the 
following dimensions hold : 5 feet long, 4 feet deep, 9 feet 
wide? 

5 feet = 60 inches. 
4 feet = 48 inches. 
9 feet = 108 inches. 

Then 60 X 48 X 108 = 311040 cubic inches. 

231)311040(1346.49 gallons. 
231 

800 
693 



1074 
924 



1500 
1386 

1140 
924 



2160 
2079 



1346, etc., gallons. Ans. 

Example. How many gallons of oil will a tank 9 feet 
3 inches long, 4 feet 5 inches wide, and 6 feet 2 inches 
deep, hold? 



236 A HAND BOOK FOR MECHANICS 

9 feet 3 inches = 111 inches. 
4 feet 5 inches = 53 inches. 
6 feet 2 inches = 74 inches. 

Then 111 X 53 X 74 = 435342 cubic inches. 

231)435342(1884.597 gallons. 
231 
2043 

1848 
1954 
1848 



1062 
924 



1380 
1155 

2250 

2079 



1710 
1617 

93 

1884, etc., gallons. Ans. 

Example. What weight of water is there in a tank 
of the dimensions of the preceding example? 

As 1 gallon of water weighs 8J pounds, 1884 gallons 
will weigh 1884 X 8J. 

1884 
8.3 
5652 
15072 



15637.2 
15637, etc., lbs. = weight. Ans. 

The best process, perhaps, for determining the weight 
of the contents of tanks is to multiply the cubic contents 
in feet by the weight of a cubic foot of the contents. 



CIRCULAR MEASURE 237 

Example: What will be the, weight of fresh water in a 
tank, when full, of the following dimensions: 5 feet long, 4 
feet wide, 9 feet deep? 

5 X 4 X 9 = 180 cubic feet. 

Then as 1 cubic foot of fresh water weighs 62.5 pounds, 
180 X 62.5 = weight of 180 cubic feet of fresh water. 

Thus 180 

62.5 
900 
360 
1080 



11250.0 = weight. 

11,250.0 1b. Ans. 

Example. A tank 5 feet long, 4 feet wide, and 9 feet 
deep, will hold how many pounds of sea water? 
One cubic foot of sea water = 64.3 lb. 

5 X 4 X 9 = 180 cubic feet. 

Then 180 X 64.3 = 11.574.0 lb. weight. 

Example. What will a tank of linseed oil of the 

following dimensions weigh: 5 feet deep, 4 feet wide, and 

9 feet long? w , . r 

5X4X9 = 180 cubic feet. 

Specific gravity of linseed oil = . 9347 

62.5 

46735 
18694 
56082 



Weight of cubic foot of oil = 58.41875 

180 X 58.418 = 10,515.240 lb. Ans. 

Example. How many pounds of linseed oil will a 
tank 5 feet 2 inches by 4 feet 5 inches, and 2 feet 5 inches 
deep, hold? 



238 A HAND BOOK FOR MECHANICS 

Reduce all dimensions to inches. We then have 
62 X 53 X 29 = 95 . 294 cubic inches 

and 95.294 cubic inches = 55.14, etc., cubic feet. 
Then 55 . 14 etc. 

58.418 weight of cubic foot of linseed oil 
44112 
5514 
22056 
44112 
27570 



3221 . 16852 lb. 

3221 lb. Ans. 

Measurements of Circular Tanks 

Example. How many cubic feet of space in a cylin- 
drical tank whose diameter is 25 inches and height 12 
inches? (Fig. 53.) 

This is the same as finding the volume of a cylinder. 

25 diam. 
25 

125 .7854 

50 625 

625 diam. squared. 39270 

15708 
47124 



490 . 8750 area of base. 
12 inches high. 
5890 . 5000 cubic inches in volume. 
Then 5890.5000 ^ 1728 = 3.408 cubic feet. Ans. 

Example. If the above cylinder is filled with linseed 
oil, what will be its weight? 



CIRCULAR MEASURE 



239 



1 cubic foot of linseed oil weighs 58.418 pounds. 

58.418 
3.408 
467344 
233672 
175254 



199.088544 



199 lb. Ans. 




Fig. 53 

How many gallons will the above cylinder hold? (Fig. 
53.) 

To find the Capacity of Cylindrical Vessels in 

Gallons. 

Rule. Multiply the area in inches by the height in 
inches and divide product by 231. 

25 X 25 X .7584 X 12 = 5890.5000 area in inches. 

231)5890.5000(25.5 gallons. Ans. 
462 

1270 
1155 

1155 
1155 



240 A HAND BOOK FOR MECHANICS 

Example. How many cubic feet, and what will be 
the weight of, and how many gallons of water will a 
tank of the following dimensions hold: diameter 43 
inches, and height 24 inches? 

43 X 43 X .7854 X 24 = 34852.9104 cubic inches. 
Then 34852.9104 - 1728 - 20.111 cubic feet. 
Weight equals 20.111 X 62.5 = 1256.937 lb. 

Now we can find the number of gallons contained in 
the tank by either dividing 34852.9104 cubic inches, 
which is its contents in cubic inches, by 231, because 
there are 231 cubic inches in a gallon. 

Thus 34852.9104 -- 231 = 150.878 gallons. 
(Almost 151 gallons.) 

Or we can find the number of gallons contained in the 
tank by multiplying its cubic contents in feet (20.111) 
by 1\, because there are 1\ gallons in one cubic foot. 

Thus 20.111 

7.5 



100555 
140777 



150.8325 gallons. Arts. 

Example. How many gallons of water will a tank 
having the following dimensions hold: base diameter 5 
feet, top diameter 3 feet, and height, 4 feet? (Fig. 54.) 

This is the first example of a frustum-shaped tank we 
have had, and the process for determining the cubic 
contents of a frustum was explained in Mensuration. 

First find cubic contents in inches, then reduce cubic 
contents in inches to cubic feet, and multiply by 7.5; 
product will be required answer. 



CIRCULAR MEASURE 



241 



36 X 36 X .7854 = 1017.8784 area in inches of 

top diameter. 
And 60 X 60 X .7854 = 2827.4400 area in inches of 

bottom diameter. 
3845.3184 = sum of areas of 

both bases. 
16 = i of perpendicular 

height in inches. 
230719104 
38453184 



61525.0944 = area in cubic inches. 
61525.0944 ^ 1728 = 35.604 cubic feet. Ans. 




Fig. 54 



Example. How many gallons will the above tank 
hold? 

35.604 X 7.5 = 267.03 gallons. 



242 



A HAND BOOK FOR MECHANICS 



Examples Explaining how to Calculate the Weight 
of Different Materials and the Weight of the 
Different Parts of a Machine 

Examples coming under this head will now offer very 
little difficulty in solving. 

To find the weight of any body, we proceed first by 
finding its area, and having obtained the area the weight 
is found by multiplying the weight of a cubic foot of the 
material by the area in feet. Or, if the area is given in 
inches, the weight will be found by multiplying the area 
in inches by the weight of a cubic inch of the material. 

Example. What is the weight of a cast-iron plate of 
the following dimensions? 4' wide 6" thick and 8' long. 
(Fig. 55.) 




Fig. 55 

4 feet = 48 inches. 
6 inches = 6 inches. 
8 feet = 96 inches. 
Then 48 X 6 X 96 = 27648 cubic inches area 
The specific gravity of cast iron being 7.110 the weight, 
of a cubic foot of cast iron will equal 

7.110 X 62.5 = 444.3750 1b. 



CIRCULAR MEASURE 



243 



And since a cubic foot weighs 444.3750 pounds, a cubic 
inch will equal 444.3750 + 1728 = .257 lb. 

Then, since one cubic inch equals .257, the weight of 
27648 cubic inches is 27648 X .257 = 7105.536 lb. 

Answer: About 7105^ pounds will be the weight of the 
plate. 

Example. What is the weight of a cast-iron plate of 
the following dimensions? (Fig. 56.) 




Fig. 56 



3 feet 6 inches = 42 inches. 
5 feet = 60 inches. 
4 feet = 48 inches. 

First find the area in cubic inches of whole piece. 

Then find the area of the whole, and subtract the latter 
from the former, and the answer will be the area in cubic 
inches of iron in the plate. 

Then multiply the area thus obtained by .257 (the 
weight of a cubic inch of cast iron) and the product will 
equal the weight. Thus: 



244 A HAND BOOK FOR MECHANICS 

60" X 48" X 42" = 120. 960 cu. inches area of plate. 

24" X 24" X .7854 X 42" = 19.000 cu. inches area of hole. 

Difference 101.960 number of cu. inches. 

in plate. 
.257 weight of cu. inches of 

713720 CaSt ir ° n ' 

509800 
203920 



26103.720 weight of plate. 
26.103 pounds equals weight of plate. Arts. 

Example. What will be the weight of a cast-iron 
column 20 feet long, outside diameter 14 inches, and 
inside diameter 8 inches? (Fig. 57.) 

63971 lb. Ans. 




Ti 



I *- 



1/ 



Fig. 57 



14" X 14" X .7854 X 240" = 36945 etc., area in in. of out- 
side diameter. 
8" X 8" X .7854 X 240" = 12063 etc., area in inches of 

inside diameter. 
Difference 24882 = number of cu. inches 

of iron. 
Then 24882 X .257 = 6394.674 lb. 
About 6397^ pounds weight of column. Ans. 

Example. What is the weight of a wrought-iron 



CIRCULAR MEASURE 245 

shaft having two flanges, one of which is 3f inches thick 
and 1 foot 5 inches diameter, the other 2\ inches thick 
and 1 foot 4^ inches diameter, the shaft is 5^ inches 
diameter and 14 feet 2 inches over all? (Fig. 58.) 

1475.327544 etc. lb. Ans. 

17 X 17 X .7854 X 3.75 = 851.1772 area of flange 1. 
16.5 X 16.5 X .7854 X 2.5 = 534.5628 area of flange 2. 
170" - 3. 75" - 2.5" = 163.75" (163|") length of shaft. 
5.5 X 5.5 X .7854 X 163.75 = area of shaft part. 

5.5 

5.5 8511772 

275 5345628 

275 38904298 



3025 5276 . 1698 = whole area in inches. 

.7854 ^28 = weight of cubic inch of 

12100 420093584 wrought iron. 
15125 105523396 
24200 1475.327544 lb. = weight. 
21175 



23.758350 
163.75 

118791750 

166308450 

71275050 
142550100 
23758350 
3890.42981250 = area in inches of shaft part. 

In the above example we first find the area of flange 1 
then the area of flange 2. We then find the area of the 
whole shaft minus the two flanges. The three areas are 
then added together and the sum thus found is the whole 
area of shaft, which multiplied by the weight of a cubic 



246 



A HAND BOOK FOR MECHANICS 



inch of wrought iron will give the whole weight of the 
shaft. 

i 



r> 






1 










' 
























bal 



X 



> 



f-m- 



m.i. 



Fig. 58 



The weight of a cubic inch of wrought iron is obtained 
by multiplying 7.690, the specific gravity of wrought 
iron, by 62.5, the weight of a cubic foot of water (fresh). 
The product is the weight of a cubic foot of wrought 
icon, which, divided by 1728, gives the weight of a cubic 
inch, .285. 

Example. A circular brass plate is 6 feet 5 inches in 
diameter, 4 inches thick, and has five 4-inch holes in it. 
(Fig. 59.) What will it weigh? 

5512.565 lb. Ans. 
77 X 77 X .7854 X 4 = 18626.5464 area of whole plate in 

cubic inches. 
4 X 4 X .7854 X 4 = ' 50.2656 area of one hole in 

cubic inches. 



251.3280 = area of five holes in 
cubic inches. 



CIRCULAR MEASURE 



247 



Then 18626.5464 - 251.3280 = 18375.2184 etc., whole 
area minus area of holes. 







Fig. 59 



Specific gravity of brass equals 8.384. 
Then brass per cubic foot equals 8.384 X 62.5 = 524 lb. 
Hence the weight per cubic inch is 524 -v- 1728 = .3 lb. 
Then the weight of 18375.2184 cubic inches is 

18375.2184 X .3 = 5512.565 lb. Arts. 

Example. What is the weight of a brass cylinder 
whose inside diameter is 10 inches, the brass being \ inch 
thick and the cylinder 3 feet 5 inches long? (Fig. 60.) 

202.8 etc. lb. - Arts. 

10 X 10 X .7854 X 41 = 3220.1400 area of inside diam. 
1 +;. 10 + J = 11 outside diameter. 

Then 11 X 11 X .7854 X 41 = 3896.3694 area of out- 
side diameter. 

Then area of outside diameter minus area of inside 
diameter equals area of metal. 

Thus 3896.3694 - 3220.1400 = 676, etc. 



248 



A HAXD BOOK FOR MECHANICS 



Then as one cubic inch of brass weighs .3 of a pound, 



676 cubic inches weighs 



676 
.3 

202.8 lb. Arts. 




Fig. GO 



Example. What is the weight of a brass cylinder 
whose length is 45 inches, outer diameter 12 inches, and 
thickness f inch? (Fig. 61.) 

357.84 lb. Ans. 




mzzzzzzzzzzzzzzzm 



'////////////////7m 



45 



^ 



Fig. 61 



CIRCULAR MEASURE 



249 



Inner diameter equals 12 — f — f = 10.5 
12 X 12 X .7854 X 45 = 5089.39 cu. in. of metal in out- 
side diameter. 

10.5 X 10.5 X .7854 X 45 = 3896.56 eu. in. of metal in 

inside diameter. 

Then 5089.39 - 3896.56 = 1192.83 cubic inches of 
metal. 

Hence 1192.83 X .3 - 357.84 lb. weight. 

Example. Find the weight of the rim of a cast-iron 
fly-wheel whose outer diameter is 9 feet 6 inches, and 
inner diameter 9 feet 3 inches, and 12 inches wide. (A cubic 
inch of cast iron equals .257 of a pound.) (Fig. 62.) 

1634.9 lb. Ans. 




Fig. G2 

114 X 114 X .7854 X 12 = 122484.700 cu. in. of metal 

in outside diam. 
Ill X 111 X .7854 X 12 - 116122.960 cu. in. of metal 

inside diam. 
Then 122484.700 minus 116122.960 equals 6361.740 
cubic inches of iron in rim. 

Hence 6361.740 X .257 = 1634.9 etc. lb., weight. Ans. 



250 A HAND BOOK FOR MECHANICS 

Example. Find the weight of the rim of a fly-wheel 
whose outer diameter is 10 feet 9 inches, and inner 
diameter 8 feet 2 inches,- and width S inches. 

11363.2 etc. lb. Ans. 

Example. What is the weight of a cast-iron cap of 
the following dimensions, if a cubic inch of cast iron 
weighs .257 of a pound? (Fig. 63.) 

First find the weight of the flanges A and B. then 
find the weight of cap, and the sum of both weights 
equals whole weight. 



Flanges. 


Outside Dimension. 


6 X 2 X 36 X 2 
2 

12 
36 


20 X 9 X 36 
9 

180 
36 


72 
36 


10S0 
540 


432 

2 = flanges. 


6480 
4032 


864 area in eu. in of flanges. 


2448 area in eu. in. cap, 




Inside Dimensions. 




16 X 7 X 36 
7 
112 
36 




672 
336 



4032 

Then 2448 + 864 X .257 = weight. 

S64 



3312 
.257 
231S4 
16560 
6624 
851 . 184 lb. Ans. (About 852 lbs.) 
Note. — Width of flange from A to C = 6" 



TABLE OF WEIGHTS 



251 




k ft" fr . /fe" j^g 



Fig. 63 
TABLE OF WEIGHTS 

(The approximate weight of a cubic inch of different 

metals) 
1 cubic inch of zinc weighs .252 lb. 
1 cubic inch cast iron weighs .26 lb. 
1 cubic inch wrought iron weighs .28 lb. 
1 cubic inch of steel weighs .288 lb. 
1 cubic inch brass weighs .3 lb. 
1 cubic inch copper weighs .32 lb. 
1 cubic inch lead weighs .41 lb. 
1 cubic inch tin weighs .262 lb. 

(The approximate weight of a cubic foot of different 

metals) 
1 cubic foot zinc weighs 428 lb. 13 oz. 
1 cubic foot cast iron weighs 450 lb. 7 oz. 
1 cubic foot wrought iron weighs 483 lb. 6 oz. 
1 cubic foot steel weighs 487 lb. 12 oz. 
1 cubic foot brass weighs 543 lb. 12 oz. 
1 cubic foot copper weighs 547 lb. 4 oz. 
1 cubic foot lead weighs 709 lb. 8 oz. 
1 cubic foot tin weighs 455 lb. 11 oz. 



252 



A HAND BOOK FOR MECHANICS 



Weight in pounds of a square foot of different metals, 
in thickness varying by 1-16 of an inch. 



Name of Metals 



5C 

go 43 
co o 

C C 

.2 a 

43 — 


+3 

43 . 

c £ 


c 
c 

1— 1 

GO 

o3 


CO 
+3 

m 


CO 

a 
a 
o 
Q 


CO 

a 


GO 
GO 

03 

pq 


"3 

CO 

a 

o 




T3 

CO 

*1 


i 

1 6 


2.3 


2.3 


2.5 


2.9 


2.3 


2.6 


2.7 


2.4 


3.7 


1 


5.0 


4.7 


5.1 


5.8 


4.7 


5.3 


5.5 


4.8 


7.4 


3 
1 6 


7.5 


7.0 


7.6 


8.7 


7.0 


8.2 


8.2 


7.2 


11.2 


1 


10.0 


9.4 


10.2 


11.6 


9.4 


11.0 


10.9 


9.6 


14.9 


5 
1 6 


12.5 


11.7 


12.8 


14.5 


11.7 


13.7 


13.7 


12.0 


18.6 


3 
8 


15.0 


14.1 


15.3 


17.2 


14.0 


16.4 


16.4 


14.4 


22.3 


T7T 


17.5 


16.4 


17.9 


20.0 


16.4 


19.2 


19.1 


16.8 


26.0 


1 

2 


20.0 


18.7 


20.4 


22.9 


18.7 


21.9 


21.9 


19.3 


29.7 


9 
1 6 


22.5 


21.1 


25.0 


25.7 


21.1 


24.6 


24.6 


21.7 


33.4 


5 
g 


25.0 


23.5 


25.5 


28.6 


23.4 


27.4 


27.3 


24.1 


37.1 


11 

1 6 


27.5 


25.8 


28.1 


31.4 


25.7 


30.1 


30.0 


26.5 


40.9 


3 
4 


30.0 


28.1 


30.6 


34.3 


28.1 


32.9 


32.8 


28.9 


44.6 


13 
1 6 


32.5 


30.5 


33.2 


37.2 


30.4 


35.6 


35.0 


31.3 


48.3 


"8 


35.0 


32.8 


35.7 


40.0 


32.8 


38.3 


38.2 


33.7 


52.0 


1 5 

] 6 


37.5 


35.2 


38.3 


42.9 


35.1 


41.2 


41.0 


36.1 


55.7 


1 


40.0 


37.5 


40.8 


45.8 


37.5 


43.9 


43.7 


38.5 


59.4 



Note. — The weight per square foot to any gage can 
easily be obtained from, the above table by multiplying 
the weight of a square foot of the metal ONE inch 
thick by the thickness of the gage in inches or parts of 
an inch. 



PART V 

THE PRIMARY OR SIMPLE MACHINES 



THE MECHANICAL POWERS 

Although the combinations of motions are many, they 
are all related to, and spring from, only two primary 
principles, namely, the principle of the " Lever" and the 
principle of the " Inclined Plane." 

The lever is the fundamental base of all circular or 
angular action, that is to say, the lever is the primary 
element of all action or motion which may be about an 
axis or center, and the inclined plane is the fundamental 
base of all rectilinear action, that is to say, the inclined 
plane is the primary element of all straight-line action 
or motion. 

The primary, or simple machines, sometimes called the 
mechanical powers, are seven in number, called: 

1. The lever. 

2. The inclined plane. 

3. The cord. 

4. The pulley. 

5. The wheel and axle. 

6. The wedge. 

7. The screw. 

The first three, that is, the lever, the inclined plane, 
and the cord, are simple elements, that is, each is an 
individual element within itself, capable of performing 
some function. And the other ones are made up of these 
three; that is, the pulley, the wheel and axle, the wedge, 
and the screw are combinations of the first three powers, 
and as the law of each one of the powers is explained in 
detail, the relation of the "pulley" and the " wheel" 

255 



256 



A HAND BOOK FOR MECHANICS 



and "axle" to the lever, and the relation of the wedge 
and screw to the inclined plane, will be clearly seen. 



THE SIMPLE MACHINES 
The Lever 

By lever is meant a stiff bar of any shape, either 
straight or bent, which is free to turn about a fixed 
point called the fulcrum (which means a stationary prop 
or support). When acted upon by two forces in opposite 
directions, one force acts as a power and the other as a 
load. 

Therefore, in the law of the lever three points must be 
considered: 

First. The fulcrum or point, about which the bar 
turns. 

Second. The point where the power is applied. 

Third. The point where the load, resistance, or weight 
is applied. 

Levers are divided into three classes. 

In the first class the fulcrum (F) is between the power 
(P) and the weight (W). (Fig. 64.) 




Fig. 64 

P = Power. 
F = Fulcrum. 
W = Weight, in all figures. 



THE SIMPLE MACHINES 



257 



In the second class the weight is between the power 
and the fulcrum. (Fig. 65.) 



u) 




— 1 



Fig. 65 



In the third class the power is between the fulcrum 
and the weight. (Fig. 66.) 




D) 




ii 'I 


il 



u) 



Fig. 66 

In all three classes, the law of the lever is the same. 
To Find the Power 

Rule. Multiply the weight by its distance from the 
fulcrum and divide by the distance of the power from 
the fulcrum. 

Example. How much power, acting at a point on a 
lever arm, 16 inches from the fulcrum, will balance a 
weight of 5 pounds located at a point 8 inches from the 
fulcrum? (Fig. 67.) 



258 



A HAND BOOK FOR MECHANICS 



8X5 



40 



2J lb. pressure. 



16 16 

That is, the pressure exerted by a weight of 2\ pounds 
16 inches from the fulcrum, would exactly counterbalance 
a weight of 5 pounds 8 inches from the fulcrum. 



tp 




-— tf*-ifc 


1 -/</'.- 


^ 










IS6 


Hill 


m- >i 



Fig. 67 

Example. How much power will have to be exerted 
on a crowbar to raise a stone which weighs 200 pounds, 
if the crowbar is 5 feet long and the fulcrum is placed at 
a point 18 inches from the end nearest the weight? 
(Fig. 68.) 




Fig. 68 

Here the leverages are 18 and 42 
200 X 18 3600 



42 



42 



= 85f lb. 



THE SIMPLE MACHINES 



259 



That is, a pressure power, or force, of 85f pounds would 
with the above lever, counterbalance - a weight of 200 
pounds. 

Example. How much power, or force, will be exerted 
to raise a block of stone which weighs 50 pounds, by a 
lever of the second order, if the stone is placed 9 inches 
from the fulcrum and the lever measures 5 feet? (Fig. 69.) 



50 



w 



-5' 



Fig. 69 



-H 



51 X 9 ' 459 



60 



60 



7-i^ lb 

' 20 1U - 



Ans. 



This answer will appear confusing to some. For such, 
it will be clearly understood, if they will think of the 
50-pound weight as being placed at the point B of the 
lever, then to raise the weight from that point it would 
take a force or power equal to 50 pounds. Now remove 
the weight from the point B to a point C, which is located 
in the middle of the lever, and a force, or power, of 25 
pounds exerted at the point B would be sufficient to 
raise it. From this the following may be deduced. 

The nearer the weight is to the point A, or fulcrum, 
the less the power required at the point B to raise it. 

Example. What power, exerted at a point located 
3 inches from the fulcrum, will balance a weight of 
120 pounds placed 3 feet from the fulcrum? 



260 A HAND BOOK FOR MECHANICS 

This is a lever of the third class. (Fig. 70.) 




120 X 36 



Fig. 70 

4320 
~3~ 



1440 lb. pressure. 



To Find the Weight 

Rule. Multiply the power by its distance from the 
fulcrum and divide by the distance of the weight from 
the fulcrum. 

Example. What weight located 36 inches from the 
fulcrum will be counterbalanced by a force of 50 pounds 
acting 6 inches from the fulcrum? (Fig. 71.) 



6 X 50 300 



36 



36 



= 84- Ans. 



Then a power pressure, or weight of 8 J pounds placed 
at a distance of 36 inches from the fulcrum will balance 
a weight of 50 pounds placed 6 inches from the ful- 
crum. 



THE SIMPLE MACHINES 



261 



61 



zv 



36"- 



50 



~V 



\0 



Fig. 71 

Example. Let A, B, C, be a bent lever with a force 
equal to 40 pounds applied at A on the long arm, the 
length A, B, is 16 inches, and the length of the short 
arm, B, C, is 4 inches. What weight at C can be lifted? 
(Fig. 72.) 




Fro. 72 



262 



A HAND BOOK FOR MECHANICS 



40 X 16 640 



= 160 lb. Am. 



Example. Let A, B, C, be a bent lever with a force 
equal to 80 pounds applied at A on the short arm, the 
length A, B, is 12 inches, and the length B, C, is 96 
inches. What weight at C can be lifted? (Fig. 73.) 

80 X 12 960 
96 = 9b" 



10 lb. Arts. 




Fig. 73 



Example. If a force of 112 pounds is applied 12 
inches from the fulcrum, what weight at A, which is 
36 inches from the fulcrum, can be lifted? (Fig. 74.) 




Fig. 74 



THE SIMPLE MACHINES 
112 X 12 1344 



263 



36 



36 



= 37+ lb. Am. 



To Find the Distance of the Power from the 

Fulcrum 

Rule. Multiply the weight by its distance from the 
fulcrum and divide by the power. 

Example. How far from the fulcrum will a power of 
50 pounds have to be placed to raise a weight of 300 
pounds, which is 9 inches, from the fulcrum? (Fig. 75.) 




Fig. 75 



300 X 9 2700 



50 



50 



= 54 inches. Arts. 



Example. Let A, B, C, 
be a bent lever with a 
power or force equal to 40 
pounds applied at A. 
How long will A, B, have 
to be to lift a weight of 
150 pounds if A, C, equals 
7 inches? (Fig. 76.) 




Fig. 76 



264 



A HAND BOOK FOR MECHANICS 



150 X 7 1050 



40 



40 



26| inches. 



Ans. 



To Find the Distance of the Weight from the 

Fulcrum 

Rule. Multiply the power by its distance from the 
fulcrum and divide by the weight. 

Example. How far from the fulcrum will a weight 
of 500 pounds balance a power of 300 pounds placed 
72 inches from the fulcrum? (Fig. 77.) 




Fig. 77 



72 X 300 21600 



500 



500 



= 43.2 inches. Ans. 



Weights Between Two Supports 

If a weight is attached to a beam which rests upon 
two supports, the beam acts as a lever of the second class, 
and the part of the whole weight carried by either sup- 
port may be found by considering one support as the 
power and the other as the fulcrum. If the weight rests 
in the middle of the beam it is obvious that each support 
will carry half the burden. But if, as shown in Fig. 78, 
the load or weight is placed a distance equal to one 



THE SIMPLE MACHINES 



265 



quarter the length of the beam from A, the support A 
will bear three quarters the weight and the support B 
one quarter. 




3 



Fig. 78 



Example. A beam 16 feet long is resting on two 
supports, A and B, there is a weight of 4 tons hanging 
5 feet from A. What share of this weight will be sup- 
ported by A and B respectively? (Fig. 79.) 




F-G. 79 

Support A holds 2f tons. 
Support B holds 1^ tons. 



Arts. 



By the principle of the lever, the shorter arm must 
bear the greater weight and the longer arm the smaller 
weight. 



266 A HAND BOOK FOR MECHANICS 

This fact was clearly demonstrated by Fig. 69. 

The following figure will clearly explain the solving of 
all such problems as the above. 

Consider A, B, Fig. 80, a lever of the second class, 
16 feet long, with a weight of 4 tons placed 5 -feet from 
the fulcrum. How much power, or what pressure ex- 
erted at B, will exactly counterbalance it? 




jr*-s <y> 



Fig. 80 



5X4 20 iii a 
= — = 11 tons. Arts. 

16 16 4 

That is to say, a weight of 1^ tons arranged over a pulley 
C, as shown in figure, would exactly counterbalance a 
weight of 4 tons placed as in the above. 

Now place a support D under the lever at B, take away 
weight E letting lever rest on support D, it is obvious 
then that support D will be resisting a pressure of 1 J tons, 
or that D will be supporting a weight of 1\ tons. 

And as according to the law of the lever the shorter 
arm (in this case the distance from the fulcrum to weight) 
bears the greater weight, it is very clearly seen that the 
share of the whole weight supported at A must be 2f tons, 
because 4 — 1J ■= 2f. 

Example. A beam 19 feet long and weighing 700 
pounds is resting on two supports A and B, there is a 



THE SIMPLE MACHINES 



267 



weight of 4 tons hanging 6 feet from A and another 
weight of 8 tons hanging 6 feet from B. What share of 
these weights and beam will be supported at A and B ? 
(Fig. 81.) 




Fig. 81 



First find the share each support bears of weight 1. 

weight of weight 1, supported 



6X8 48 _ 

= — = 2.63 tons 

19 19 



at A. 

Then as A supports 2.63 tons of the weight 1, B must 
support 5.37 tons of the weight 1. 

In the same manner find the share each support bears 
of weight 2. 

6X4 24 

= — =. 1.27 tons = weight of weight 2 at B. 

19 19 s s 

Then as B supports 1.27 tons of weight 2, A must sup- 
port 2.73 tons of weight 2. And each support bears also 
half the weight of the beam equals 350 pounds. 

Therefore, A's share = 2.63 + 2.73 + 350 = 
and B's share = 5.37 + 1.27 + 350 = 
Therefore A's share equals 2.63 + 2.73 + 350; A's share 
in pounds equals 5260 + 5460 + 350 = 11,070 lb., about 
5 tons, etc. And B's share equals 5.37 -f 1.27 + 350; B's 



268 



A HAXD BOOK FOR MECHANICS 



share in pounds equals 10,740 + 2540 + 350 = 13.630 = 
6 tons 10 cwt. 20 lb. = about 6£ tons. 

The Compound Lever 

By compound lever is meant a combination of levers. 
Any system of two or more levers acting upon each other 
is called a compound lever. 

By the use of compound levers a very small force 
applied will sustain a great weight. 

And compound levers are used when and where it is 
inconvenient to use a single lever having a long arm. 

Figure 82 shows a system of levers called a compound 
lever. 



A 

\&i — // 



* 



5@7 



•J3 










Fig. 32 



With such a system of levers as the above, where the 
short arm of one works on the long arm of the other 
lever, a very small power applied at B would lift or bal- 



ance a great weight at A. 



To Fixd the Power 

To find the power exerted at B in a combination of 
levers arranged as in Fig. 82. to balance a weight of 440 
pounds suspended from A. 

Rule. Multiply the weight by continued product of 
short arms and divide by continued product of long arms. 

Thus in the above Fig. 82, the short arms are 3 inches, 



THE SIMPLE MACHINES 269 

3 inches, and 2 inches, and the long arms are 11 inches, 

9 inches, and 8 inches. 

Therefore the power required at B to balance a load 
of 440 pounds at A is 

440 X 3 X 3 X 2 7920 in A 

= = 10. Arts. 

11X9X8 792 

That is, a power of 10 pounds exerted at B will balance a 
weight of 440 pounds suspended from A, or 10 pounds 
suspended from B will balance a weight of 440 pounds sus- 
pended from A. 

To Find the Weight 

Rule. Multiply the continued product of the long 
arms by the power, and divide by the continued product 
of the short arms. 

Example. If there be such a combination of levers 
as represented in Fig. 82, with long arms of 11 inches, 9 
inches, and 13 inches, and short arms of 2 inches, 3 inches, 
and 2 inches, what weight will be balanced by a power of 

10 pounds? 

w • U4r 10 X 11 X 9 X 13 12870 in7Q - „ 

Weight = == = 1072.5 lb. 

2X3X2 12 

Then a pressure or power of 10 pounds exerted at A 

would balance a weight of 1072| pounds suspended 

from B. 

The Pulley 

The pulley is a wheel over which a cord, chain, or band 
is passed in order to transmit the force applied to the 
cord, chain, or band in another direction. 

The pulley is really a combination of the cord and a 
wheel, or of a cord and a number of wheels, or of a number 
of cords and a number of wheels. 

The pulley or wheel is introduced only to reduce fric- 



270 



A HAND BOOK FOR MECHANICS 



tion, and the usefulness of the pulley is dependent wholly 
upon the cord. 

Pulleys are said to be fixed or movable according as 
their blocks are fixed or movable. 

Figure S3 shows a fixed pulley. 

In this pulley the block A is fixed or stationary, and 
the wheel C, D. turns within it. 




O O 



Fig. S3 

In Fig. 84 is shown a single movable pulley. In this 
pulley the block A is movable and the wheel C turns 

within it. 




Fig. 84 



THE SIMPLE MACHINES 



271 



There is no power or mechanical advantages gained 
by a single rope acting over one or more single fixed 
pulleys, but there is often great advantage gained by 
their use, because, as the force applied to the cord is 
transmitted in another direction, a fixed pulley or com- 
bination of fixed pulleys enables us to change the direc- 
tion of the force. 

Thus it is easier for a man by the use of a fixed pulley 
to hoist a weight to a loft than it would be for him to 
carry the weight upwards over a flight of stairs. 



Work Done by the Pulley 

The pulley may be considered as a continuous series 
of levers with equal arms on one fulcrum or axis. 

Figure 85 represents a fixed pulley which corresponds 



t 



u * 



A 



\ 




--C 



6 6 



Fig. 85 

to a lever of the first class, the center line A, B, C, repre- 
senting the elements of the lever in which B, the center, 
may be looked upon as being the fulcrum, and A, B, and 
A, C, the lever arms. 

It is here obvious that the single fixed pulley changes 
the direction of the force only, without modifying the 



272 



A HAND BOOK FOR MECHANICS 



intensity of the power. That is, a weight of 100 pounds 
suspended from A would exactly counterbalance a weight 
of 100 pounds suspended from C, because A, B, and B, C, 
are equal, but a force applied to the cord at A, causing 
it to move in a downward direction, would cause the 
cord at C to move in an upward direction. 



Combination of Pulleys 

Movable pulleys are generally used in combination 
with fixed pulleys. Fig. 86 shows a combination of 
one fixed pulley with one movable pulley. 




Fig. 86 

The figure shows a cord attached to a beam at A, then 
led around a single movable pulley, and then up over a 
single fixed pulley; the free end B is where the power is 
applied, and attached to the movable pulley is a weight 
= W. 

In such an arrangement it is evident that the weight W 
is supported equally by the two parts C, D, of the cord 
which passes around the movable pulley, that is to say, 
the whole weight is supported by the cord and each of 
the two singles C and D of the cord bears one half the 
weight ; and since no power is gained by the fixed pulley K 
in this combination it is seen the force at B required to 
balance the weight W will be half the weight. That is, 



THE SIMPLE MACHINES 



273 



50 pounds suspended from B will balance a weight of 1 00 
pounds suspended from the movable pulley, because each 
part of the cord C and D (called each single of the rope) 
supports one half the weight. In the combinations of 
pulleys in common use, several fixed pulleys are con- 
tained in one block, and in the other block are an equal 
number of movable ones. 

Figure 87 shows a combination of this kind having 
two fixed pulleys in the upper block 
and two movable pulleys in the lower 
one. In all such combinations of 
pulleys, one continuous cord passes 
through the system. The weight here 
when the lift begins is supported by 4 
singles of the rope, and therefore the 
power required to balance the weight 
in this combination will be one fourth 
of the weight. That is, in this system 
the tension of the weight is equally 
distributed among the four parts of 
the cord which sustains the weight, 
hence a power applied at P will coun- 
terbalance four times its weight at W. 

From this, then, the following may 
be deduced: 



To Find the Power 

Rule. Divide the weight by the 
number of singles of rope. 




Fig. 87 



Example. A piece of iron weighing 9 tons is lifted 
by a pair of blocks of two sheaves each, the rope is fast- 
ened to the upper block. What power is required to 
lift it? (Fig. 87.) 



274 A HAND BOOK FOR MECHANICS 

9 

7 = 21 tons. Ans. 
4 

Example. A cylinder cover weighing 1100 pounds is 
lifted by a pair of blocks of two sheaves each, the rope 
is fastened to the upper block. What power is required 
to perform the work? (Fig. 87.) 

-^ = 275 lb. Ans. 
4 

The answers to above examples only tell the amount 
of power required at B, Fig. 86, and at P, Fig. 87, 
to evenly balance weights suspended as shown in figures. 
If it be required to raise the weights, additional power 
must be applied to overcome friction. 

Example. A pair of double blocks are used to lift 
a ton weight. What power must be applied if 20 per cent 
be lost by friction? 

Power required without friction = = 500 lb. 

With friction: If the power were 100 pounds (a cwt.), 
then the weight lifted would be only 80 pounds, as 20 
per cent is lost. 

Hence, as 80 : 100 :: 500: power required. 

100 



8.0)50000(625. Ans. 
48 
20 
16 



40 
40 



Example. A pair of double blocks are used to raise 
a weight of 2 tons. What must be the power applied at 



THE SIMPLE MACHINES 275 

the free end of the rope, if 20 per cent be lost by fric- 
tion? 

Weight in pounds = 2000 X 2 = 4000 lb. 

Power required without friction, = 1000. 

With friction, 80: 100 :: 1000: power required. 

100 
8.0)100000(1250. Ans. 
8_ 
20 
16 

40 

40 



The Strength of a Rope 

The strength of a rope depends upon its area or cir- 
cumference; the circumference called the girth being gen- 
erally used. The rule is that the girth or circumference 
squared and divided by 24 is equal to the weight that 
may be lifted expressed as tons. The divisor 24 varies 
with the quality of the rope. 

Example. It will be safe to lift how many tons with 
a rope whose girth is 14 inches? 

14 2 -r- 24 = Si tons. Ans. 

Example. What weight may be lifted by a rope 
2J inches in girth? 2J = 2.5. 

2.5 2 -i- 24 = 260 lb. Ans. 

Example K. What weight may be safely lifted by 
a rope whose girth is 3 inches? 

3 2 -T- 24 - | of a ton = 750 lb. 



276 A HAXD BOOK FOR MECHANICS 

To Find the Breaking Stress of a Rope 

Rule. — Multiply the girth squared by .28, the product 
is the breaking stress in tons. 

Example L. — What is the breaking stress of a rope 
3 inches in girth? 

3 2 X .28 = 2.52 tons = 4500 lb. 

Note that Example K shows that it is only safe for a 
rope of 3 inches girth to lift 750 pounds, and that Example 
L shows that a rope whose girth is 3 inches will not break 
till a weight of 2.52 tons or 4500 pounds be applied. 

In practice the working load for round rope should 
not exceed a seventh of the ultimate strength and for 
flat rope one ninth. Therefore, if the ultimate strength 
of a rope whose girth is 3 inches equals 2.52 tons or 4500 
pounds it would not be safe to put a working load on 
that rope exceeding one seventh of 4500 pounds, 642 
pounds, although in practice often such a rope carries 
700 to 800 pounds load. 

Example. — What is the breaking stress of a rope 
4J inches in girth? 

4.5 2 X .28 = 7.23 tons. Ans. 

The Wheel and .axle 

The wheel and axle may be likened to a couple of 
pulleys of different diameters united on one axis of which 
the larger is the wheel and the smaller the axle. 

As shown in Fig. 88 the combination consists of a 
wheel, or drum A mounted upon an axle B. 

The power is applied at the cord wrapped around the 
wheel A. and the weight or resistance suspends from a 
cord wrapped around the axle B. 



THE SIMPLE MACHINES 



277 



The wheel and axle is really a lever of the first class 
and may be treated as such. In which, as shown in 




Fig. 89, the center line A, C, represents the elements of 
the lever, D being the fulcrum about which the lever 
turns. 




Fig. 89 



The radius D, F, of the wheel will be the leverage of the 
force applied, or power, and the radius E, D, of the axle the 
leverage of the weight. 



278 



A HAND BOOK FOR MECHANICS 



From this, then, we may deduce the following rules: 
To find the power: multiply the weight by the radius 
of the axle and divide by the radius of the wheel. 

Example. What power will be required to lift a 
weight of 300 pounds if the diameter of the drum or wheel 
is 12 inches and the diameter of the axle 6 inches. 
(Fig. 90.) 




Fig. 90 
Radius of axle == 3 inches; radius of wheel = 6 inches. 

Ans. 



3 X 300 900 , Kn 
= = 150. 

6 6 



That is, a pressure of 150 pounds at R will balance a 
weight of 300 at K. 

To find the weight: Multiply the power by the radius 
of the wheel and divide by the radius of the axle. 



THE SIMPLE MACHINES 



279 



Example. What weight will be balanced by a power 
of 75 pounds if the diameter of the wheel is 10 inches 
and the diameter of the axle 5 inches? (Fig. 91.) 

Radius of axle = 2.5 inches; 
radius of wheel = 5 inches. 



75 X 5 375 



= 150 lb. 



2.5 2.5 

Then a weight of 150 pounds 
at A will be balanced by a pres- 
sure of 75 at B. 

Example. The diameter of 
a steering wheel is 4 feet and the 
barrel is 14 inches diameter.. 
What resistance will be over- 
come if a man applies a force 
equal to 200 pounds. (Fig. 92.) 

4 feet = 48 inches diameter, 
or 24 inches radius of wheel; 14 
inches diameter = 7 inches radius 
of axle. 




Fig. 91 




Fig. 92 



280 



A HAND BOOK FOR MECHANICS 



Then 20 ° X 2i , 687 lb. 



The Windlass 

The windlass is a combination of the wheel and axle 
in which a crank answers the purpose of the wheel. 

The windlass consists of an axle A, B, and a crank 
C, D, E, Fig. 93, by means of which the axle A, B, is 
turned. 




Fig. 93 



As shown in figure, the crank is an arm standing per- 
pendicular to the axle. 

The part 2 of the crank is called the crank arm, and 
the part 1 is called the crank handle. 

The windlass is used for raising heavy weights and is 
operated by applying the power to the crank handle 1, 
the weight to be raised being made fast to the cord, which 
is fast to and wraps around the axle as same is made to 
revolve. 



THE SIMPLE MACHINES 281 

Example. ■ What weight will be raised by a windlass, 
if a pressure of 150 pounds be applied to the crank handle? 
From center of axle to center of crank handle 15" and 
axle 10" diameter. 

l*i<J$2_ 450 11, An.. 

5 

In the last example suppose 10 per cent of the force 
applied be lost through friction and other causes, what 
would have been the weight lifted? 

10 per cent of 450 = 450 

10 
45X10 

Then 450 - 45 = 405 lb. Arts. 

In other words, as 10 per cent, that is one tenth is loss, 
nine tenths must be left. 

Hence, -^ of 450 = 405 lb. Arts. 

In the examples given, the diameter of the rope has 
not been considered because ropes of the sizes required 
to lift such weights would be too small in diameter to 
consider. If, however, the rope used is very thick, then 
the leverage must be measured from the center of the 
axle to the center of the rope. 

Example. In a windlass, the center of crank handle 
is 10 inches from center of axle and from center of axle to 
center of rope is 6J inches. What weight will a pressure 
of 150 pounds raise, 10 per cent being lost in friction? 

10X150 1500 . A „ .., ,, , , , . 

■ = = 240 lb. without loss due to friction. 

6.25 6.25 

10% of 240 = 24. 
Then 240 - 24 = 216 lb. Ans. 



282 



A HAND BOOK FOR MECHANICS 



The Differential Windlass 

The differential windlass sometimes called the Chinese 
wheel and axle, as shown in Fig. 94, differs from the com- 
mon windlass, Fig. 93, in having an axle formed by two 
drums A and B of different diameters. A cord is at- 
tached to the larger cylinder and several times wrapped 
around it, then passes under a movable pulley C, and then 
is wrapped in an opposite direction around the smaller 
cylinder B. 

The power is applied at the crank arm as in the common 
windlass, but the weight is suspended from the movable 
pulley C. 





1 




h 


1 3> 


\ 






^wfca 


1 


' | 




Ik ^^^ 














A' 











iiiiii. limn 



Fig. 94 



With this arrangement of windlass, when the handle is 
turned so as to wind up rope on the cylinder A it is at 
the same time unwound from the cylinder B, and at each 
revolution of the crank handle the rope is shortened only 
by the difference in the circumference of the cylinders. 
The greater the difference in the circumferences, the 
quicker the weight will move, and the less the difference 



THE SIMPLE MACHINES 283 

the slower it will move, and by having them nearly equal 
the weight moves very slowly and great power is gained. 

Example. In a differential windlass the diameters of 
the drums are 1 foot and f of a foot, the length of the 
crank arm is 2 feet 3 inches and the power applied to 
the crank handle is equal to 90 pounds. What weight 
will be lifted? 

Rule. Multiply the length of crank arm in feet by 
the power, and divide by half the difference in feet of the 
the radii of the drums. 

Thus 2 feet 3 inches = 2\ feet = 2.25 feet, length of 
arm in feet. 

Then 2.25 X 90 = 202.50 -f- J difference in feet of radii 
= Arts. 

As the diameters of the drums are 1 foot and f feet 
(=9 inches) the radius of the larger drum will equal 6 
inches and the radius of the smaller drum equals 4| inches. 

Then the difference between the radii is 6 — 4.5 = 1.5 
inches difference between radii of drums. 

Then J of 1.5 = f = .75 of an inch. 

Now reduce .75 of an inch, which is \ the difference of 
the radii in inches, to the decimal of a foot. 

Thus 12). 75 (.0625 of a foot. 

72 

30 
24 



60 
60 



202 ^ 
Then, = 3240 lb. Ans. 

.0625 



284 



A HAND BOOK FOR MECHANICS 




The Inclined Plane 

The inclined plane is a flat surface inclined to the 
horizon. Any plane which forms with a horizontal plane 
any angle whatever excepting a right angle may be con- 
sidered an inclined plane. 

In Fig. 95 the line A. then, is an inclined plane because 

it is a surface inclining 
to the horizontal line B. 
The inclined plane is 
used for raising weights, 
and by referring to Fig. 
95 it will be seen that 
less power would be re- 
quired to raise a given 
weight by the use of the sloping path formed by line A 
than would be required if surface A approached nearer 
to the perpendicular as shown in Fig. 96. 

The less the height of the plane in proportion to its 
length, or the less the angle of inclination, the greater the 
mechanical effect ; and the greater the height of the plane 
in proportion to its length the less 
will be the mechanical effect. 

The following explanation will 
more clearly explain this ; 

When a body rests on a horizon- 
tal plane, say for example on a 
table, the action of gravity tend- 
ing to draw it down is completely 
counteracted by the resistance of 
the plane on which it rests and the body remains at 
rest. 

It is not so, however, when a body is placed on an in- 
clined plane. In this case the force of gravity acts upon 




Fig. 96 



THE SIMPLE MACHINES 



285 



the body in two ways: one perpendicular to the plane, 
which tends to force the weight through it, and the other 
parallel to the plane, which tends to draw the weight 
along parallel to it. 

The action of the force of gravity in the first place is 
counteracted by the resistance of the plane, whilst in the 
second place the body meeting no resistance in a line 
parallel to the plane it will be forced along in that direc- 
tion by the force of gravity acting parallel to the plane. 
And it is evident that the nearer the plane approaches to 




Fig. 97 



a horizontal surface, the greater will be the portion of 
weight supported by the surface. Let the plane be 
elevated toward the perpendicular and it will support 
less and less of the weight, till when it reaches an exact 
perpendicular it will be supporting no part of the weight 
at all. 

Fig. 97 will demonstrate this still more clearly. 

In the figure is shown two weights of unequal sizes, 
A and B, connected by a cord C which passes over a pulley 



286 A HAND BOOK FOR MECHANICS 

D. The larger weight A is resting on the inclined surface 
and is balanced by the smaller weight B. It is evident 
that the nearer the inclined surface on which the larger 
weight A is resting approaches the horizontal line E the 
greater will be the portion of the weight A supported by 
it, and consequently the less the weight at B will be re- 
quired to counterbalance A. 

Also, the nearer the inclined surface on which A is 
resting approaches a perpendicular the greater the weight 
at B will be required to counterbalance A. 

From this we learn that the small weight B will coun- 
terbalance a weight larger than itself which Tests on an 
inclined plane; and demonstrates the fact that by the 
use of the inclined plane a given weight can be raised by 
a power which is less than the weight itself. 

Rules for Calculating Work Done by the Inclined 

Plane 

To find the power: Multiply the weight by the height 
of the plane and divide by the length. 

Perpendicular height of plane 

p _ W v/ £ 2 i 

length of plane 

Example. What force is necessary to keep a weight 
of 100 pounds stationary on an inclined plane whose per- 
pendicular height is 4 feet and length of incline 14 feet, 
supposing there is no friction? 

Thus P = i^- 4 = 4™ 28f lb. 
14 14 

p = Power or force. 

Example. The length of an inclined plane is 19 feet, 
the perpendicular height 6 feet. What power will be re- 



THE SIMPLE MACHINES 



287 



quired to sustain a weight of 250 pounds, supposing 
there is no friction? (Fig. 98.) 




Fig. 98 



ThusP = 



250 X 6 1500 



19 



19 



78.9 lb. Ans. 



To find the weight : Multiply the power by the length 
of the plane and divide by the height. 



W = 



PX L 
H 



W = Weight 
P = Power 

L = Length of plane 
H = Height 

Example. The length of an inclined plane is 15 feet, 
the perpendicular height 7 feet. What weight will a power 
of 78 pounds sustain, supposing there is no friction? 



78 X 15 



= 167.1 lb. Ans. 



Example. The length of an inclined plane is 15 feet, 
the perpendicular height 3 feet. What weight will be sus- 
tained by a power of 78 pounds? 



288 



A HAND BOOK FOR MECHANICS 
I^li 5 = I 170 = 390 lb. Ans. 



The Wedge 

The form of wedge in general use is the double wedge 
represented in Fig. 99, which may be likened to a pair of 




Fig. 99 

inclined planes joined together back to -back along the 
dotted line A, B. 

There is, however, another form of wedge used called 
the single wedge, represented in Fig. 100, which may be 
likened to the inclined plane. Wedges are used where it is 




Fig. 100 

required to exert a great force in a small space. No 
accurate estimate can be made of the work done by 
wedges as they are ordinarily used, but calculations worked 
out according to the following rules will give an approx- 
imate idea of the amount of work which may be performed 
by them. 



THE SIMPLE MACHINES 



289 



To Find the Power 

Rule. Multiply weight by thickness of wedge and 
divide by length of wedge. 

Example. A wedge 20 inches long and 3 inches 
thick is employed to lift a weight of 200 pounds. What 
power must be exerted? (Fig. 101.) 



to 

« 

1 




Fig. 101 



200 X 3 

~~ 20 



600 
20 



30 lb. pressure 



Example. If a wedge 12 inches long and 3 inches 
thick is employed to raise a weight of 1000 pounds, what 
pressure must be exerted? 



P = 



1000 X 3 3000 



12 



12 



= 250 lb. Arts. 



To Find the Weight 

Rule. Multiply the power by the length of the 
wedge and divide by the thickness. 

P = power 
P X L L = length 

~~ T T = thickness 

W = weight 



Weight = 



290 



A HAND BOOK FOR MECHANICS 



Example. If a wedge is 15 inches long and 2 inches 
thick, what weight will it raise if the power applied is 
100 pounds? 



W = 



100 X 15 1500 



= 750 lbs. 



2 -2 

Example. If a double wedge 12 inches long 4 inches 

thick splits a block when it is 
driven in with a force of 200 
pounds, the pressure exerted 
upon the block by the wedge is 
how many pounds? (Fig. 102.) 




W 



200 X 12 2400 



= 600 lb. 



The Screw 

Fig - 102 The screw is essentially an in- 

clined plane wrapped around a cylinder, as may be seen 
by taking a triangular piece of paper and winding it 
around a cylindrical barrel. 

Take, for example, the inclined plane A, B, C (Fig. 
103), and bend it in circular form resting on its base line 




Fig. 103 



A, B; by so doing a helical plane 1, 2 will be formed as 
shown in Fig. 104, which figure shows the inclined plane 
A, B, C (Fig. 103) curled around the cylinder D. 

Here, then, it is seen that as the incline winds around 



THE SIMPLE MACHINES 



291 



the cylinder D it assumes a spiral shape more accurately 
called helical. The screw, then, is really a combination 
of inclined planes, because it consists of a solid barrel A 




Fig. 104 

(Fig. 105) enveloped by a spiral projection called the 
thread B, which is nothing more than an inclined plane 
wound around a solid barrel. 




Fig. 105 



It is not clearly determined how far the mechanical 
powers were known to the ancients. Without doubt they 



292 A HAND BOOK FOR MECHANICS 

had a knowledge of the lever, the wheel and the axle, and 
the pulley, and it would seem because of the great en- 
gineering feats of construction and building performed 
by the early Egyptians, that they must have had a knowl- 
edge of the principles of the inclined plane; this is only 
surmised and not an established fact, but it seems very 
probable, in order to have moved the huge blocks of stone 
of which the pyramids are built, that they must have had 
an acquaintance with the inclined plane. Just when 
the inclined plane was first used in the form of a screw 
is not known. 

The earliest authentic mention we have of the screw 
being used in mechanics is in the writings of Archimedes. 
This philosopher and mechanician, about 236 B.C., in- 
vented a pumping screw or spiral cylinder for raising 
water. The device is now known as " Archimedes' 
screw," and is one of the most ancient contrivances for 
raising water. 

This screw consists of a metallic tube, wound in a 
spiral form around a solid cylinder or shaft, which is 
made to revolve by turning a handle. When placed at 
the proper angle, the water, as the handle is turned, will 
continue to flow into those parts of the tube that are 
brought successively below the shaft, till gradually it 
will be discharged at the top. (Fig. 106.) 

Apparently Archimedes did not, however, understand 
the inclined plane as now used, for he makes no direct 
mention of it in any of his writings, and there is no posi- 
tive evidence to show that it was included in the 
knowledge of mechanics possessed by the Romans. 

The inclined plane seems to have been invented by 
Galileo in the sixteenth century, because several writers 
of that time allude to the wedge and the screw, showing 
that a knowledge of these powers formed a part of the 



THE SIMPLE MACHINES 



293 



revival of physical science in which Galileo took a most 
prominent part at that time, even if he did not fully 
inspire it. However, Stevinnes, a mechanician of Hol- 
land, was the first to fully explain in a treatise the theory 
of the power of the inclined plane and screw. 

The first use of the screw was' in the screw jack, an 
appliance used for raising heavy weights. 




Fig. 106 



The various modifications and applications of this 
power belong to the era of mechanical discovery of the 
present century. 



Work Done by the Screw 

Screws may be either single-threaded or double- 
threaded. If we assume that a screw consists of a cylin- 
der with a coil which forms the thread wound around it, 
we may easily define a double screw as a cylinder with 
two coils parallel to one another wound around it. The 



294 A HAND BOOK FOR MECHANICS 

screw works into a solid, fitted with a thread to receive 
it, called the nut. The nut may be fixed or movable: 
when fixed the screw turns within it, and when the nut 
is movable the screw is fixed, the nut being made to turn 
upon it. 

Motion is imparted to the screw or the nut (whichever 
the case may be) by means of a lever at the extremity 
of which the power is applied. The longer the lever and 
the less the distance between the threads, the greater 
will be the force exerted at the point of resistance. 

Screws are said to be right-handed or left-handed. A 
right-handed screw is one which, passing through a fixed 
nut and turned toward the right hand, will advance into 
the nut. A left-handed screw is one which will, by turn- 
ing it in a direction toward the left hand, be made ad- 
vance into a fixed nut. 

A screw in one revolution will advance into a fixed nut 
a distance equal to its pitch, or a distance equal to the 
distance between the centers of two successive threads. 
That is to say, if the distance between the centers of two 
successive threads be I inch, the screw during one com- 
plete turn will advance into a fixed nut that distance. 

The distance between the centers of two successive 
threads is always called the pitch of the screw. 

The following rules give an approximate knowledge 
of work performed by the screw: 

To Find the Power 

Rule. Multiply the weight in pounds by the pitch 
and divide by the circumference of the circle described 
in inches by the handle employed to turn either the screw 
or nut one revolution. 

Example. If the pitch be f inches, and the lever 3 



THE SIMPLE MACHINES 



295 



feet long, what power at A will be required to move a 
weight of 5 tons? (Fig. 107.) 



[ 






-«■ 



•a* 



A^ 



T 



I — ,~ ! 



n 



Zbunu* 



I 



I i^/^C* 



Fig. 107 



^ 



W* 



IK 



J 



P = Power 

Twice length of lever arm = 6 feet — 72 inches. 
Then 72 X 3.1416 = 226.1952 inches circumference 
of circle described by revolution of lever. 

5 tons = 2000 X 5 = 10000 lb. 



Thus P = 



10000 X f 
226.1972 



33.1 lb. 



296 A HAND BOOK FOR MECHANICS 

Example. If the pitch of a screw be \ inch, and the 
lever be 3 feet long, what force would move 5 tons? 

Twice length of lever = 6 feet = 72 inches; 72 X 3.1416 
= 226.1952 inches circumference of circle described by 
lever. 

Weight in pounds = 2000 X 5 = 10000 lb. 

10000 X .25 2500 , , „ 

P = == = 11 lbs. 

226.1952 226.1952 

It will be noticed from the above examples, that the- 
oretically there is no limit to the power of the screw, for 
the distance between the threads may be made as small 
as you please, and the lever may be made as long as you 
please; the limits will be mechanical, viz.: the strength 
necessary to be given to the threads and the space neces- 
sary to be given to the lever. 

To Find the Weight 

Rule. Multiply power by the circumference in 
inches of the circle described, by the handle or lever 
employed to either turn the screw or the nut, and divide 
by the pitch. 

Example. If the distance between the threads be 
\ inch (= .25), and a force of 100 pounds be applied to 
the end of a lever 2 feet in length, what weight will be 
moved by the screw? 

Thus: Twice length of lever = 4 feet — 48 inches. 

100 lb. X 48 X 3.1416 15079.6800 6031glb 
.25 ~ .25 

The rule mostly used in practice for determining the 
weight that can be lifted by a screw, or the tension that 
can be given a bolt or stud by using a wrench to the nut, 



THE SIMPLE MACHINES 297 

is the distance moved by the power or the hand in making 
one complete revolution of the wrench, divided by the 
pitch of the screw and the quotient multiplied by the 
power applied. 

From this deduct 70 to 75 per cent for friction. 

All machines, no matter how complicated or what the 
work is they perform, are combinations of these simple 
mechanical devices, or primary machines, arranged to- 
gether and combined so as to give such direction and 
velocity to the motion that the required work is per- 
formed. 



PART VI 

STRENGTH OF MATERIALS AND QUESTIONS 
RELATING TO STRESS 



STRENGTH OF MATERIALS AND QUESTIONS RELATING 

TO STRESS 

The mechanic, to work intelligently, should be ac- 
quainted with the strength of the different materials 
used in construction. He should also be familiar with 
the strains, due to the various forces, any of the great 
variety of materials used will stand before final rupture. 
And it is therefore necessary that he have a knowl- 
edge of those laws which govern the rules for calculating 
the extent of the strain due to the various forces to be 
dealt with, and know what to allow as a safe margin of 
resistance to carry all such forces, and thereby make the 
required provisions for the strength of parts. 

The external forces applied to materials tending to 
cause their rupture or alteration of form are called stresses. 
These are five in number, called: 

1. Tensial stress. 

2. Compressive stress. 

3. Transverse stress. 

4. Torsional stress. ' 

5. Shearing stress. 

A force applied in the line of fiber having a tendency 
to lengthen the body is called the tensial or breaking 
stress. 

As shown in Fig. 108, the beam A has a tendency to 
lengthen because of the force exerted upon it by the 
weight B, and the resistance of A due to the load B to 
alter its form is the stress, — in this case the tensial 
stress. 

301 



302 



A HAND BOOK FOR MECHANICS 



A force which pushes the parts together, tending to 
shorten or crush the body, is called the compressive stress. 

As shown in Fig. 109 the weight A exerts a stress, 
called compressive stress, upon the column B, tending 
to crush it. 




A force applied to a body in such manner as to cause 
it to twist is called the torsional stress. As shown in Fig. 
110, by applying a power to the lever A in such manner 



STRENGTH OF MATERIALS 



303 



that it would cause the beam B, which is anchored at C, 
to twist. 




Fig. 109 




Fig. 110 

If a beam is resting on two supports and a weight is 
placed on the beam between the supports, it is called a 
transverse stress. (Fig. 111.) 



304 



A HAND BOOK FOR MECHANICS 



If the beam be fixed at one end and loaded at the other, 
the stress is a bending stress. (Fig. 112.) 




Fig. Ill 

Lastly, if the stress is such that it acts in a way that it 
is trying to cut through a body it is called a shearing 
stress. As shown in Fig. 113, where the cutters A, B, 
are cutting through the bar C. 



6 



Fig. 112 



The immediate result of stress is strain. 

Every load which acts on a structure produces a 
change of form, which is called the strain due to the load. 
The strain may be either a vanishing or elastic change of 
form; that is, one which disappears when the load is re- 



STRENGTH OF MATERIALS 



305 



moved, or it may be a permanent change or set, which 
remains after the load is removed. Strain, then, is the 
alteration in shape as the result of stress. 

And in designing structures, care must be taken that 
all parts will have the required strength, so that under 
the greatest straining action there will be no sensible 
permanent change in form. 

The stress required to strain, change the form of or 
rupture the various materials in use is usually deter- 
mined by an apparatus called the Testing Machine, of 



L 




3£l 



Fig. 113 



which there are many different forms, and it may be here 
pointed out that the differently constructed machines 
of this class give widely different results, there being a 
great variation in the data indicated by them represent- 
ing the standard resistances of the various materials 
tested. 

Therefore, the careful mechanic, rather than depend 
upon the data given by the different authorities as being 
correct in representing the strength of the different 
materials, is obliged to himself test and determine the 
strength of the materials he proposes to use, and will 



306 



A HAND BOOK FOR MECHANICS 



regard given data rather as approximate and subject to 
change than as precedents to be adopted and blindly 
followed. 

It will be the object of this chapter to explain the rules 
for determining the strain of the approximate stresses due 
to the action of the different loads, to show by tables 
and otherwise the approximate strength of the various 
materials used, and how to make comparative safe pro- 
vision for sustaining all those forces which enter as fac- 
tors in construction of every nature. 

As cast iron, wrought iron, steel and wood, are the ma- 
terials mostly used in machine construction, special note 
had better be made of their various strengths, so that 
intelligent judgment may be used when these materials are 
employed to construct the parts of such sizes that they 
will safely stand the stresses due to the action of the dif- 
ferent loads, and thereby prevent straining and rupture. 
The stresses to which constructions and parts of con- 
structions are mainly subjected to are the Tensial, Com- 
pressive, and Shearing stresses. 

Stress is usually measured in pounds per square inch 

of sectional area. That is, 
the ultimate strength of 
- materials is usually deter- 
mined by the number of 
pounds a square inch of 
sectional area will undergo 
before final rupture. 

For instance, if a 
wrought-iron bar is 2 
inches by 2 inches in sec- 
tion. (Fig. 113.) Thus 
the whole area of section is 4 square inches, and it is 
obvious that if the ultimate tensial strength of wrought 




T 
i 



Fig. 113 



STRENGTH OF MATERIALS 307 

iron, for instance, is 52,000 pounds per square inch of 
section, a bar of wrought iron whose sectional area is 
4 square inches would be caused to break by a power 
of 52,000 X 4 = 208,000 pounds. 

The following tables are approximate of the ultimate 
strength and elastic strength of different materials com- 
monly used. 

Ultimate Strength 

By ultimate strength is meant the greatest possible 
load any piece of material will carry before fracture is 
produced. In other words, the smallest load which will 
cause the rupture of a piece of material is called the 
ultimate strength of that piece; that is, the stress in 
pounds per square inch which the piece can sustain just 
before rupture takes place. 

Elastic Strength 

All materials are more or less elastic and will undergo, 
when put under great stress, change of form which may 
be vanishing or permanent as before stated. When the 
load is such that it has a permanent influence upon the 
piece and sets it so that it will not return to its original 
form when the load is removed, the limit of elasticity of 
the material has been reached (which is the elastic 
strength), and beyond this limit the strain increases 
faster than the stress until rupture is produced. 



308 



A HAXD BOOK FOR MECHANICS 



TABLES 



TABLE NO. 1 



Ultimate and Elastic Strength of Materials in 
Pounds per Square Inch 



NAME OF 

MATERIALS 



Cast Iron. 



ULTIMATE OR 
BREAKING STRENGTH 



Tension. „^??? „ Shearing 
pression & 



Wrought Iron . 



Soft Steel En- 
hardened . . . . 



Soft Steel hard- 
ened 

Cast Steel Un- 
tempered. . . . 



Cast Steel Tem- 
pered. . 

Copper 

Brass 

Gun Metal 



10.800 90.000 20.000) 

to 
67.000 
52.000 52.000 

to 
67.000 



60.000 150.000 

to 
100.000 



"0 . 000 



Phosphor 
Bronze. 



120.000 120.000 120.000 

84.000 84.000 84.000 

to to to 

150.000 150.000 150.000 

100.000 150.000 70.000 

33.000 58.000 58.000 

17.500 10.500 10.500 

23.000 23.000 23.000 

to to to 

52.000 52.000 52.0001 



58.000 58.000 58.000 



ELASTIC STRENGTH 



Tension ^^„?^" Shearing 
pression & 



10.500 21.000 7.900 
24.000 24.000 20.000 



35.000 35.000 



26.000 



70.500 70.500 53.000 
80.000 80.000 74.000 



S4.000 90.000 145.000 

4 . 300 3 .900 2 . 900 

6.950 6.950 5.200 

6.200 6.200 -4.150 



19.700 19.700 14.500 



STRENGTH OF MATERIALS 



309 



TABLE NO. 2 

Table Showing Safe Working Stress in Pounds 
per Square Inch 

ordinary working stress 



NAME OF 
MATERIALS 



Cast Iron 

Wrought Iron . 
Soft Steel Un 

tempered . . . 
Cast Steel Un 

tempered. . . 

Copper 

Brass 

Gun Metal .... 
Phosphor 

Bronze. 



SAFE LIMITS IN POUNDS 
PER SQUARE INCH 



Tension 



3.600 
10.400 

17.700 

36.000 
3.600 
3.600 
3.120 

9.870 



Com- 
pression 



10.400 
10.400 

17.700 

36 . 000 
3.120 
3.600 
3.120 

9.870 



Shearing 



THEORETICAL LIMIT OF 
STRESS 



Tension 



2.700 
7.800 

13.000 

20.000 

2.300 
2.700 
2.400 

7.380 



5.250 
17.280 

24.000 

52.000 
9.900 
5.250 

10.800 

17.400 



Com- 
pression 



28.500 
15.000 

24.000 

52.000 

17.400 

3.150 

10.800 

17.400 



Tension 

Com- 
pression 



3.000 
9.000 

13.000 

36.000 
5.500 
3.000 
6.000 

9.700 



As shown by foregoing tables, 1 and 2, there is a great 
margin of difference between the ultimate strength and 
the safe working limit allowed for the load or weight that 
the various metals will safely carry before straining. 

This allowance is made because it is impossible to de- 
termine all the forces which produce straining action; 
therefore, to insure the safety of a structure, we have to 
make all parts sufficiently strong to support, not only 
the aggregate amount of straining action due to the 
forces which are taken into consideration, but also to 
take care of all unforeseen contingencies due to neglected 
causes of straining action. 

Hence it is necessary that the mechanic be sufficiently 
informed so that he be able to determine the approximate 



310 



A HAND BOOK FOR MECHANICS 



stress on a structure, and also be able to determine the 
necessary strength of the different parts to prevent strain 
and thereby insure absolute safety. And although by 
practical experience and familiarity he perhaps gains the 
best knowledge of the intensity of straining actions due 
to the different forces, it is customary in practice, when 
determining the straining actions on a structure, to be 
more or less guided by various factors of safety. 

As just stated, it is impossible to take into considera- 
tion all the forces which produce a straining effect upon 
a structure. However, by multiplying those forces 
which are known to exist by a factor of safety, a rough 
and comparatively safe allowance is made for the straining 
action which is exerted by the unseen forces. And in 
this way is obtained the approximate stress on the struc- 
ture due to all causes. 

And the sizes and strength of the various materials to 
be used to carry those various stresses may be deter- 
mined accordingly. 

Following is a table of factors of safety, which are 
generally adopted in practice for various materials, under 
dead and varying or live load, and for machines subjected 
to sudden and frequent strains of short duration, known 
as shocks. 

FACTORS OF SAFETY 



Name of Materials 


Dead Load 


Live Load 


Materials 

Subject to 

Shocks 


Cast Iron 


6 
4 
4 
5 
8 
15 


10-15 

6 

7 

8-10 
10 
25 


15-20 


Wrought Iron 


12 


Steel 


15 


Copper 


10-15 


Timber 


15 


Masonry and Brickwork . . . 


30 



STRENGTH OF MATERIALS 311 

It will be noticed that the factor of safety is less for 
dead loads than for live or varying loads, and greatest 
when the structure is subjected to shock. 

From which we learn that the stress due to the different 
loads varies; therefore the nature of the load to be car- 
ried must be carefully considered and provisions accord- 
ingly made to carry them. 

It being impossible to correctly determine the intensity 
of a load all the parts must be made to resist a much 
greater load than will be brought to bear on them at any 
time; therefore the expected load is multiplied by one of 
the various factors of safety. And this factor of safety 
varies according to the nature of the load, and con- 
sequently for the same materials under the influence of 
different loads, a greater or less factor of safety must be 
used. 

Tensial Stress 

If a bar of iron 1 inch square is torn asunder by a stress 
of 22 tons, what will be required to break a bar 3f" 
square? That is, if a square bar whose sides measure 1" 
is torn asunder by a stress of 22 tons, what will be 
required to break a square bar whose sides measure 3| " ? 



3.75 2000 


= 1 ton. 


3.75 22 




1875 4000 




2625 4000 




1125 44000 stress in 


14.0625 sq. inches in section of bar. 


p ounds 


44000 


required 


562500 


to break 


562500 


bar 1 inch 


618750.0000 lbs. to break the bar. Arts. 


square. 



Example. If a bar of cast iron 2 inches square is torn 



312 A HAND BOOK FOR MECHANICS 

asunder by a stress of 100,000 pounds, what will be 
required to break a bar of 3J inches square? 

3.5 

3.5 



175 
105 

12.25 sq. inches in section of bar to be broken. 

2 
2 

4 sq. inches in section of bar which is broken. 

Then if a force of 100,000 pounds is required to break a 
bar of cast iron of 4 square inches sectional area, how 
much force will be required to break a cast-iron bar of 
12.25 square inches sectional area more? 

Thus, 4 : 12.25 :: 100,000: Ans. 

12.25 



500000 
200000 
200000 
100000 
4 )1225000.00 

306250 lb. to break the bar. Ans. 

Example. If a bar of wrought iron 2 inches in diam- 
eter is torn asunder by a weight of 70 tons, what is its 
breaking stress? 

It will be remembered that stress is measured in pounds 
per square inch of section; and that breaking stress 
means the stress in pounds per square inch of section 
required to break any material. 

Hence, the question of the above example is to deter- 
mine the stress in pounds required to rupture a square 
inch of (in this instance) wrought iron. 



STRENGTH OF MATERIALS 313 

Thus 2 
2 
4 = square of diameter of bar. 

Then, .7854 

4 

3.1416 sq. inches in section of bar. 

Then if 70 tons is the breaking stress of 3.1416 square 
inches, what is the breaking stress of 1 square inch? 

Thus, 3.1416 : 1 ::70: Arts. 

1 

3.1416)70.0000(22 tons. Arts. 
62832 

71600 
62832 

That is, each square inch of section will stand before 
breaking a force of 22 tons, or 70 -r- (.7854 X 2 2 ) = 22.28 
tons per square inch of section. Ans. 

Example. A piece of plate 9 inches long, 4 inches 
wide, and f inches thick is tested in line of its length and 
breaks at 72 tons. What is its breaking stress? 

Here the section is 4 X f = 3 square inches. 

And the breaking stress = 72 tons -~ 3 = 24 tons per 
square inch of its section. Ans. 



Compressive Stress 

Example. If a solid casting 4 inches diameter is 
crushed by a weight of 450 tons, what is the crushing 
strength of this specimen of cast iron? 

Thus: 



314 A HAND BOOK FOR MECHANICS 

Area of section = .7854 X 4 2 = 12.5664 square inches. 
Stress = 450 + 12.5664 = 35 tons. 
35 tons per square inch of its section. Am. 

Example. If a cast-iron column 5 inches by 5 inches 
holds up a weight of 25 tons, what is the stress in pounds 
per square inch on the column? 

5 
_5 

25 square inches in section of column. 
2000 = 1 ton 

25 
10000 
4000 

50000 = weight in pounds supported by the column. 
Then stress in pounds per square inch of sectional area 
- 50000 + 25 = 2000 lb. Ans. 

Example. If four solid cast-iron columns, each 4 
inches square, hold up a cistern weighing 46 tons, whose 
inside dimensions are 12 feet long, 8 feet wide, and 6 feet 
deep, what is the stress per square inch on these columns 
when the cistern is one half full of water? (Fig. 114.) 

First find weight of water in cistern when full, which 
divide by 2 to obtain weight in pounds when half full. 
Add to this the weight of cistern when empty. Divide 
the sum obtained by the total sectional area of the col- 
ums (4 of them), and the quotient 1562 is the required 
answer. 

Cubic contents of cistern when full of water = 12 X 8 
X 6 = 576 cubic feet. 

1 cubic foot of water weighs 62.5 lb. 

Weight of water in cistern = 576 X 62.5 ^ 2 = 18,000 
lb. 



STRENGTH OF MATERIALS 



315 



Total weight on columns in pounds = 92,000 (46 tons) 
and 18,000 = 100,000 lb. 

Total sectional area of columns -4X4X4 = G4 
square inches. 

Stress per square inch = 100,000 - 64 = 1562 lb. Ans. 




XJ^ 



Fig. 114 
Shearing Stress 

In cutting through a bar we cut through its section: 
then in shearing stress always find the sectional area. 

The shearing stress of wrought iron is usually con- 
sidered to be 52,000 pounds (23 tons) per square inch. 

That is, a stress of 52,000 pounds is required to cut 
through every square inch of sectional area. 

Example. What shearing stress will cut through a 
three-quarter inch rivet? 



316 



A HAND BOOK FOR MECHANICS 



.75 X .75 X .7854 = .44178750 sectional area of rivet, 
(Fig. 115.) 
Then, .44148650 X 52,000 lb. = 22.972 lb. Ans. 




Fig. 115 

The foregoing examples will serve to demonstrate the 
process for determining the magnitude of the strain car- 
ried per square inch of section which may 
be caused by the action of the 
stresses. 




various 



116 



we 



Examplks Demonstrating the Process 
for Constructing Parts of Suffi- 
cient Strength to Carry Loads of 
Known Magnitudes. 

Example. A square wrought-iron bar is 
to carry a weight of 208,000 pounds, what 
must be its sectional area? (Fig. 116.) 

Let us employ a factor of safety of 4. That 
is, we will suppose the load which is to be 
sustained will be 4 times greater than 208,000, 
which is 816,000 pounds. As the piece of 
material in question is subjected to a tensial 
find by referring to table No. 1 that the ulti- 



STRENGTH OF MATERIALS 



317 



mate tensial strength of wrought iron is 52,000 pounds 
per square inch of section. 

Therefore, to find the required area of section, divide 
the load 816,000 pounds by the ultimate strength 52,000 
pounds. 

832,000 



52,000 



16 square inches. Arts. 



Example. What must be the sectional area of a 
square cast-iron block which is to sustain a weight of 
160,000 pounds. (Fig. 117.) 




Fig. 117 

Let us employ a factor of safety, say 5 for cast iron. 
That is, we will suppose the load which is to be sustained 
will be 5 times greater than 160,000, which is 800,000, 
pounds. As the piece of material in question is subjected 
to a compressive stress, we find by referring to table No. 1 
that the ultimate compressive strength of cast iron is 
90,000 pounds per square inch of section. 

To find, then, the required area of a section, divide the 
load, 800,000 pounds by 90,000. 

Thus = 8.888 square inches. Ans. 

90,000 



318 



A HAND BOOK FOR MECHANICS 



Example. What must be the sectional area of a 
wrought-iron bolt to withstand a shearing stress of 80,000 
pounds? (Fig. 118.) 




Fig. 118 

Let us employ a factor of safety of 4 for wrought iron. 
That is, we will consider the load which is to be carried 
to be 4 times greater than 80,000 pounds that is, 320,000 
pounds. As the piece of material in question is subjected to 
a shearing stress, we find, by referring to table No. 1, the 
ultimate shearing stress of wrought iron to be 52,000 
pounds per square inch of section. 

To find, then, the area of a section capable of sustaining 
a shearing stress of 80,000 pounds, divide the load 320,000 
by 52,000. 

320,000 



Thus, 



52,000 



6.153 square inches. Ans. 



Examples Showing the Process for Determining 
the Amount of Stress on Beams 

Example. If a beam 10 feet long fixed at one end has 
suspended from it a weight of 300 pounds, what is the 
stress at the point where the beam is fixed? (Fig. 119.) 

Rule. The intensity of the force is equal to the 
force multiplied by the length of the perpendicular to 
the direction of the force from a point in which the beam 
is supposed to be fixed. 



STRENGTH OF MATERIALS 



319 



Therefore, the stress on the beam at A in pounds equals 
300 X 10 = 3000 lb. 



— "S: 



J'-V. 



.lo'. 



Q->. 



Fig. 119 



Because the length of the perpendicular from A, the 
point at which the beam is fixed to the center of the force 
300 equals 10 feet, and 300 X 10 = 3000 lb. 

Example. What is the stress at B, Fig. 120, when a 
weight of 300 pounds is suspended from the free end of 
beam? 



L — oh^=-. 




Fig. 120 



320 



A HAND BOOK FOR MECHANICS 



In this case a perpendicular drawn from the point at 
which beam is fixed to the center of line of the force is 
9 feet, therefore the stress at B = 300 X 9 = 2700 lb. 

The stress throughout the length of a beam fixed at 
one end and loaded at the other is not uniform. That 
is, the stress varies at different parts of the beam. 

The intensity of the stress is greatest at the point where 
the beam is supposed to be fixed and gradually diminishes 
in intensity toward the load. 

For instance, there is more stress on the beam at A (Fig. 
121), than at B, and the stress is greater at B than at C. 




Fig. 121 



To Find the Stress in Foot Pounds of any Sec- 
tion of a Beam 

Rule. Multiply the load by its distance from the 
section. For instance, it is required to find the stress at 
a section 5 feet from the free end of a beam from where a 
load of 100 pounds is suspended. (Fig. 122.) 

Thus 100 X 5 = 500 lb. 

That is, the intensity of the stress on the beam at a 
point (section) five feet from the load, the load being 
100 pounds, is 500 pounds. 



STRENGTH OF MATERIALS 



321 



£"- 



r-i 







Fig. 122 



Example. What is the stress on a section of a beam 
2 feet from the load, the beam being fixed at one end 
and carrying a load of 300 pounds? 

Thus, 300 X 2 = 600 foot pounds. 

The above examples fully demonstrate that the greatest 
stress is at the point where the beam is supposed to be 
fixed, and that it gradually diminishes toward the load. 

Therefore it is customary, when constructing beams of 
this nature, to make them strongest where the greatest 
stress comes. (Fig. 123.) 



6 



Fig. 123 



322 



A HAND BOOK FOR MECHANICS 



Example. If the greatest stress allowed on wrought 
iron be taken at 52,000 lbs. per square inch, what should 
be the sectional area at the wall A of a wrought-iron beam 
66 inches long, supporting a weight of 2 tons suspended 
from its free end? (Fig. 124.) 




Fig. 124 



2 tons = 4000 lb. 

4000 X 66 = 264,000 lb. stress at A. 

Let us now employ a factor of safety of 4 for wrought 
iron. That is, consider the stress 264,000 at A to be 4 
times greater = 1,056,000 lb. 

Then, 1,056,000 h- 52,000 = 20.31 square inches. Arts. 

Example. A beam 56 inches long fixed at one end 
supports a weight of 5000 pounds. What will be the 
stress on the beam at fixed end A, and what will be the 
stress at end B? (Fig. 125.) 

5000 lbs. = stress at B. 

5000 X 56 = 280,000 lb. stress at A. 

Example. A beam is 3 inches thick, 5 inches deep, 
and 56 inches long. What stress will be put on it per 



STRENGTH OF MATERIALS 



323 




Fig. 125 



square inch of its section at B, by hanging a weight of 
5000 pounds from its extremity? (Fig. 126.) 



S6- 



(A 



% 



B M^ 




Fig. 126 



Thus 3 X 5 = 15 square inches of section. 
5000 X 56 = 280,000 lb. stress at B. 
280,000 -f- 15 = 12,000 lb. per square inch. 



Arts. 



Example. If a wrought-iron beam 66 inches long 
has to carry weight of 5000 pounds, what ought the 
sectional area of the beam be at its fixed end, and what 
ought its sectional area be at its extremity? (Fig. 127.) 



324 



A HAND BOOK FOR MECHANICS 



CC? 



«3 



Fig. 127 




$040 



5000 X 66 X 4 = 1,320,000 lb. stress at B. 

5000 X 4 = 20,000 lb. stress at A. 

1,320,000 ~ 52,000 = 25.5 square inches required at B. 

20,000 ■*- 52,000 = .36 square inches required at A. 



ANSWERS 



ANSWERS 

ANSWERS TO PAGE 8 

1. Five. 

2. Twenty-five. 

3. Three hundred and forty-two. 

4. One thousand six hundred and seventy-four. 

5. Fifty-four thousand three hundred and forty. 

6. Nine hundred and sixty thousand seven hundred and eighty. 

7. Three million seven hundred and twenty-six thousand nine 
hundred and two. 

8. Fifteen million nine hundred and eight thousand six hundred 
and sixty. 

9. Three hundred and two million six hundred and seven thou- 
sand six hundred and six. 

10. Five billion three million three thousand and three. 

11. Thirty-two billion six hundred seventy-three million three 
hundred thousand three hundred. 

12. Nine hundred and ninety-nine billion nine hundred and 
ninety-nine thousand nine hundred and ninety-nine. 

ANSWERS TO PAGE 9 

1. Four thousand three hundred and sixty-four. 

2. One thousand nine hundred and twenty-seven. 

3. Nine thousand and nine. 

4. Four hundred and thirty-four thousand six hundred and 
seventy-two. 

5. Six million four hundred and ninety-seven thousand nine 
hundred and twenty-three. 

6. Fifty-three million two hundred and ninety thousand six hun- 
dred and seventy-eight. 

7. Five hundred million four hundred and ninety thousand and 
sixty-nine. 

8. Five billion eight hundred and sixty-seven million three hun- 
dred and forty thousand and sixty-eight. 

327 



328 



ANSWERS 
ANSWERS TO PAGE 10 



1. 3,725. 

2. 12,600 

3. 322,006. 



4. 6,050,020. 

5. 38,420,350. 

6. 222,835,130. 



ANSWERS TO PAGE 15 
1. 100,677. 1. .26219. 



2. 17,903,922. 

3. 2,394,627. 

4. 127,228,172. 

5. 784,666,605. 



2. 
3. 



.8552072. 
.40750202. 



ANSWERS TO PAGE 63 



1. 

2. 
3. 
4. 


8477640537. 
600238144 tf 
205761317 if. 
51549479 ff 




5. 174172502 f T 

6. 3082404. 

7. 25096125 iff 

8. 1080637 M&i 




ANSWERS 


TO 


PAGE 51 


1. 
2. 
3. 

4. 
5. 


804 | 
42066i. 
1649^ 
87235£. 
91734 \. 




7. 227307 f. 

8. 4081. 

9. 110094 fVVo. 

10. 8422115 TT- 

11. 1205456. 


6. 


113082. 








ANSWERS 


TO 


PAGE 40 


1. 
2. 
3. 


548076. 

194032332. 

2739987852. 




4. 126102012000 

5. 668968011290 



Catalogue of Scientific Publications 

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MOORE, E. C. S. New Tables for the Complete Solu- 
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52 D. VAX NOSTRAXD COMPANY'S 

SANFORD, P. G. Nitro-explosives. A Practical Treatise 

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SAUNNIER, C. Watchmaker's Handbook. A Workshop 

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SCHUMANN, F. A Manual of Heating and Ventilation 

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SCIENTIFIC PUBLICATIONS. 53 

SCHWEIZER, V. Distillation of Resins, Resinate Lakes 

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SCIENTIFIC PUBLICATIONS. 55 

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SCIENTIFIC PUBLiCAflOM 57 

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SCIENTIFIC PUBLICATIONS. 59 

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PAGE 

MECHANICAL ENGINEERING. 1 

CIVIL ENGINEERING 11 

MARINE ENGINEERING, &c. . 19 

MINING & METALLURGY 22 

COLLIERY WORKING, Sec. ... 26 

ELECTRICITY 28 

ARCHITECTURE & BUILDING. 31 
SANITATION & WATER SUP= 

PLY 35 

CARPENTRY & TIMBER 36 



PAGE 

DECORATIVE ARTS 38 

NATURAL SCIENCE 40 

CHEMICAL MANUFACTURES. 41 

INDUSTRIAL ARTS 43 

COMMERCE, TABLES, &c 49 

AGRICULTURE & GARDEN= 

ING SO 

MATHEMATICS & ARITH- 
METIC 54 

LAW & MISCELLANEOUS. ... 56 



MECHANICAL ENGINEERING, ETC. 



THE MECHANICAL HANDLING OF MATERIAL. 

A Treatise on the Handling of Material, such as Coal, Ore, Timber, etc., 
by Automatic or Semi-automatic Machinery, together with the Various 
Accessories used in the Manipulation of such Plant, and Dealing fully 
with the Handling, Storing, and Warehousing of Grain. By G. F. 
Zimmer, A.M. Inst. C.E. 528 pages. Royal 8vo, cloth, with 550 Illus- 
trations (including Folding Plates) specially prepared for the Work 

Net $10.00 
Contents: — Chapter I. Introductory. — II. Elevators. — III. Worm 
Conveyors. — IV. Push-Plate or Scraper Conveyors. — V. Trough 
Cable Conveyors. — VI. Band Conveyors. — VII. Metal Band Con- 
veyors. — VIII. Picking Belts or Tables with or without Lowering 
Ends or Shoots. IX. The Continuous Trough or Travelling Trough 
Conveyor. — X. Vibrating Trough Conveyors. — XI. Tightening Gear 
for Elevators and Conveyors, and Driving Power required for Dif- 
ferent Types of Conveyors. — XII. The Travelling or Tilting Bucket 
Conveyors. — XIII. Pneumatic Elevators and Conveyors. — XIV. Con- 
veyors DESIGNED FOR SPECIAL PURPOSES, INCLUDING THE BoLINDER TIMBER 

Conveyor, Coke Conveyors, and Casting Machines. — XV. Endless 
Chain and Rope Haulage. — XVI. Ropeways and Aerial Cableways, 
including Ropeways, Cableways, and Appliances for Coaling at Sea. — 
XVII. Unloading Appliances, including Methods of Discharging by 
means of Skips and Grabs. — XVIII. Discharging Vessels and Barges 
by means of Elevators. — XIX. Unloading by means of Specially Con- 
structed Self-emptying Boats and Barges. — XX. Unloading by means 
of Specially Constructed Self-emptying Railway Trucks. — XXI. Un- 
loading BY MEANS OF COAL TlPS. XXII. COLLIERY TlPPLERS. XXIII. 

Miscellaneous Loading and Unloading Devices. — XXIV. Automatic 
Loading Devices. — XXV. The Automatic Weighing of Material. — 
XXVI. Coaling of Railway Engines. — XXVI I. Coal-handling Plant 



2 CROSBY LOCKWOOD & SOX'S CATALOGUE. 

for Gas-works, Potter Stations, Boiler-houses, etc. — XXVIII. Floor 
and silo Warehouses for Grains and Seeds. — XXIX. Coal Stores and 
Coal Silos. — XXX. High-level Cranes. — Index. 

HOISTING MACHINERY. 

An Elementary Treatise on. Including the Elements of Crane Con- 
struction and Descriptions of the Various Types of Cranes in Use. By 
Joseph Horner, A.M.T.M.E., Author of "Pattern-Making" and other 
Works. Crown Svo, with 215 Illustrations, including Folding Plates, 
doth S3.00 

AERIAL OR WIRE=ROPE TRAMWAYS. 

Their Construction and Management. By A. J. Wallls-Tayleb, 

A.M.Inst. C.E. With SI Illust rat ions. 12mo, cloth S3. 00 

"An excellent volume, and a very good exposition of the various systems 
of rope transmission in use and gives as well not a little valuable informa- 
tion about their working, repair, and management. We can safely recom- 
mend it as a useful general treatise on the subject." — Engineer. 

MODERN MILLING MACHINES. 

Their Design. Construction, and Working. A Handbook for Practical 
Men and Engineering Students. By Joseph Horner, A.M.I.Mech.E., 
Author of ' ' Pattern -Making," etc. With 269 Illustrations. Demy 
8vo, cloth. [Just Ready.] S4.00 

TOOLS FOR ENGINEERS AND WOODWORKERS. 

A Practical Treatise including Modern Instruments of Measurement. 

By Joseph Horner. A.M.Inst .M.E., Author of "Pattern-Making," etc. 

Demy, Svo, with 456 Illustrations S3. 50 

Summary of Contents: — Introduction. — General Survey of Tools. — 
Tool Angles. — Sec. I. Chisel Group. — Chisels and Applied Forms fob 
Woodworkers. — Planes. — Hand Chisels and Applied Forms foe Metal 
Working.— Chisel- like Tools for Metal Turning. Planing, etc. — 
Shearing Action and Shearing Tools. — Sec. II. Examples of Scraping 
Tools. — Sec. III. Tools. — Relating to Chisels and Scrapers. — Saws. — 
Files. — Milling Cutters. — Boring Tools for Wood and Metal. — Taps 
and Dies. — Sec. IV. Percussive and Moulding Tools. — Punches. Ham- 
mers and Caulking Tools. — Moulding and Modelling Tools. — Miscel- 
laneous Tools. — Sec. V. Hardening. Tempering, Grinding, and Sharp- 
ening. — Sec. VI. Tools for Measurement and Test. — Standards of 
Measurement. — Squares. Surface Plates, Levels, Bevels. Protrac- 
tors, <fcc. — Surface Gauges or Scribing Blocks. — Compasses and Divi- 
ders. — Calipees, Vernier Calipers, and Related Forms. — Micrometer 
Calipers. — Depth Gauges and Rod Gauges. — Snap, Cylindrical and 
Limit Gauges. — Screw Thfead, Wire and Reference Gauges. — Indi- 
cators, etc. 

ENGINEERS' TURNING IN PRINCIPLE & PRACTICE. 

A Handbook for Working Engineers. Technical Students, and Amateurs. 

By Joseph Horner, A.M.I.Mech.E., Author of "Pattern-Making," etc. 

8vo, cloth, with 4SS Illustrations 83.50 

Summary of Contents: — Introduction. — Relations of Turnery and 
Machine Shop. — Sec. I. The Lathe. Its Work : and Tools. — Forms and 
Functions of Tools. Remarks on Turning in General. — Sec. II. Turn- 
ing Between Centres. — Centring and Driving. — Use of Steadies. — 
Examples of Turning Involving Lining-out for Centres. — Mandrel 
Work. — Sec. III. Work Supported at One End. — Face Plate Turning. 
Angle Plate Turning. — Independent Jaw Chucks. — Concentric, Uni- 
versal, Toggle, and Applied Chucks. — Sec. IV . Internal Work. — 
Drilling. Boring, and Allied Operations. — Sec. V. Screw Cuttings 
and Turret Work. — Sec. VI. Miscellaneous. — Special Work. — Meas- 
urement, Grinding. — Tool Holders. — Speed and Feeds, Tool Steel. — 
Steel Makers' Instructions. 



MECHANICAL ENGINEERING, &>c. 3 

THE MECHANICAL ENGINEERS' REFERENCE BOOK. 

For Machine and Boiler Construction. In Two Parts. Part I. Gen- 
eral, Engineering Data. Part II. Boiler Construction. With 
51 Plates and numerous Illustrations. By Nelson Foley, M.I.N.A. 
Third Edition, Revised throughout, and much Enlarged. Folio, half- 
bound In Press 

Part I: Measures. — Circumferences and Areas, &c. — Squares, 
Cubes, Fourth Powers. — Square and Cube Roots. — Surface of Tubes. 
— Reciprocals. — Logarithms. — Mensuration. — Specific Gravities and 
Weights. — Work and Power. — Heat. — Combustion. — Expansion and 
Contraction. — Expansion of Gases. — Steam. — Static Forces. — Gravi- 
tation and Attraction. — Motion and Computation of Resulting 
Forces. — Accumulated Work. — Centre and Radius of Gyration. — 
Moment of Inertia. — Centre of Oscillation. — Electricity. — Strength 
of Materials. — Elasticity. — Test Sheets of Metals. — Friction. — 
Transmission of Power. — Flow of Liquids. — Flow of Gases. — Air 
Pumps, Surface Condensers, &c. — Speed of Steamships. — Propellers. — 
Cutting Tools. — Flanges. — Copper Sheets and Tubes. — Screws, Nuts, 
Bolt Heads, &c. — Various Recipes and Miscellaneous Matter. — With 
DIAGRAMS for Valve-gear, Belting and Ropes, Discharge and Suc- 
tion Pipes, Screw Propellers, and Copper Pipes. 

Part II : Treating of Power of Boilers. — Useful Ratios — Notes 
on Construction. — Cylindrical Boiler Shells. — Circular Furnaces. 
Flat Plates. — Stays. — Girders. — Screws. — Hydraulic Tests. — Rtvet- 
ing. — Boiler Setting, Chimneys, and Mountings. — Fuels, &e. — Exam- 
ples of Boilers and Speeds of Steamships. — Nominal and Normal 
Horse Power. — With DIAGRAMS for all Boiler Calculations and 
Drawings of many Varieties of Boilers. 

THE WORKS' MANAGER'S HANDBOOK. 

Comprising Modern Rules, Tables, and Data. For Engineers, Mill- 
wrights, and Boiler Makers; Tool Makers, Machinists, and Metal 
Workers; Iron and Brass Founders, etc. By W. S. Hutton, Civil 
and Mechanical Engineer, Author of "The Practical Engineer's Hand- 
book." Sixth Edition, carefully Revised and Enlarged. 8vo, strongly 

bound $6.00 

£3^~ The Author having compiled Rules and Data for his own use in a great 
variety of modern engineering work, and having found his notes extremely use- 
ful, decided to publish them — revised to date — believing that a practical work, 
suited to the daily requirements of modern engineers, would be favorably 
received. 

"The Author treats every subject from the point of view of one who has 
collected workshop notes for application in workshop practice, rather than 
from the theoretical or literary aspect. The volume contains a great deal 
of that kind of information which is gained only by practical experience 
and is seldom written in books." — The Engineer. ' 

STEAM BOILER CONSTRUCTION. 

A Practical Handbook for Engineers, Boiler-makers, and Steam Users. 
Containing a large Collection of Rules and Data relating to Recent Prac- 
tice in the Design, Construction, and Working of all Kinds of Stationary, 
Locomotive, and Marine Steam-boilers. By Walter S. Hutton, 
Civil and Mechanical Engineer, Author of "The Works' Manager's 
Handbook," "The Practical Engineer's Handbook," &c. With up- 
wards of 500 Illustrations. Fourth Edition, carefully Revised and 

Enlarged. 8vo, over 680 pages, cloth, strongly bound $6.00 

^P~ This Work is issued in continuation of the series of handbooks 
written by the Author, viz: "The Works' Manager's Handbook" and 
"The Practical Engineer's Handbook," which are so highly appreciated 
by engineers for the practical nature of their information, and is consequently 
written in the same style as those ivorks. 

Contents: — Heat, Radiation, and Conduction, Non-conductin g 
Materials, and Coverings for Steam Boilers. — Composition, Calorific 
Power, and Evaporative Power of Fuels. — Combustion, Firing Steam 
Boilers, Products of Combustion, &c. — Chimneys for Steam Boilers,— 



4 CROSBY LOCKWOOD &■ SOX'S CATALOGUE. 

Steam Blast. — FORCE Draught.— FeEd Water. — Effect of Heat on 
Water. — Expansion of Water by Heat. — Weight of Water at Differ- 
ent Temperatures. — Convection. — Circulation. — Evaporation. — 
Properties of Saturated Steam. — Evaporative Power of Boilers. — 
Priming, etc. — Water-Heating Surfaces of Steam Boilers. — Trans- 
mission of Heat. — Smoke Tubes. — Evaporative Powers and Effi- 
ciency of Boilers. — Water Capacity a\d Steam Capacity of Boilers. — 
Eire-Grates, Eire-Bredges. and Eire-Bars. — Power of Boilers. — 
Cylindrical Shells and Furnace-Tubes of Boilers. &c. 

Tests of Materials. — Strength and Weight of Boiler-Plates. — 
Effect of Temperature on Metals. — Rivet Holes. — Rivets. — Rivet 
Joints of Steam Bollers.— Caulking. — Ends of Cylindrical Shells. — 
Stays for Boilers, ire. — Steam Generators. — Description and Pro- 
portions of Cornish. Lancashire, and Other Types of Stationary 
Boilers. — Boiler Setting. — Multitubular Boilers. — Locomotive 
Boilers. — Portable Boilers. — Marine Boilers. — Vertical Boilers. — 
Water-tube Boilers. — Superheaters. — Cost of Steam Production. — 
Furnaces for Refuse Fuels. — Destructors, <tc. 

Safety- Valves. — Steam Pipes. — Stop- Valves, and Other Mountings 
for Boilers. — Feed Pumps. — Steam Pumps. — Feed-Water Consumption. 
— Injectors. — Incrustation and Corrosion. — Feed- Water Heaters. — 
Evaporators. — Testing Boilers. — Evaporative Perfor m ances of 
Steam Boilers. Steam-Boiler Explosions, <kc. 

PLATING AND BOILER MAKING. 

A Practical Handbook for Workshop Operations. Bv Joseph G. Hor- 
ner. A.M.I.M.E. 380 pp. with 33S Illustrations. 12mo cloth.,. §3.00 
Contents : — The Trade. — Tools. — Materials. — Testing Materlals. — 
Limiting Dimensions and Weights of Materlals. — Cutting and 
Straightening Plates, &c. — Bending Plates. — Bending Angles. A:c. — 
Welding. — Flanging. — Punching. — Riveting. — Types of Riveted 
Joints. — Estimation of Lengths of Material. — The Marking Out of 
Work. — The Estimation of Weights, &e. — Machines. 

A TREATISE ON STEAM BOILERS. 

Their Strength. Construction, and Economical Working. By R. Wil- 
son, C.E. Fifth Edition. 12mo, cloth 82.50 

"The best treatise that has ever been published on steam boilers." — En- 
gineer. 

BOILER AND FACTORY CHIMNEYS. 

Their Draught-Power and Stability. With a chapter on Lightning Con- 
ductors. By Robert Wilson. A.I.C.E., Author of '"A Treatise on 
Steam Boilers," etc. 12mo, cloth SI. 50 

BOILERMAKER'S ASSISTANT 

> In Drawing, Templating, and Calculating Boiler Work, etc. By J. 

\ Courtney." Practical Boilermaker. Edited by D. K. Clark, C.E. 

Seventh Edition. 12mo, cloth .SO 

BOILERMAKER'S READY RECKONER. 

With Examples of Practical Geometry and Templating for the Ese of 
Platers, Smiths, and Riveters. Bv John Courtney. Edited bv D. 
K. Clark, M.Inst. C.E. Crown 8vo, cloth SI. 60 

BOILERMAKER'S READY RECKONER & ASSISTANT. 

With Examples of Practical Geometry and Templating for the L se of 
Platers, Smiths, and Riveters. Bv John Courtney. Edited by D. K. 
Clark. M.Inst. C.E. Fifth Edition. 480 pp., with 140 Illustrations. 
Fcap. 8vo, half-bound S3.00 

*** This Work consists of the t"-o previous-mentioned volumes. "Boiler- 
maker's Assistant - ' a^ " Boilermaker's Reapt Reckoner," found 

iOG*th& ftl Q™ Volume. 



MECHANICAL ENGINEERING, &c. 5 

STEAM BOILERS. 

Their Construction and Management. By R. Armstrong, C.E. Illus- 
trated. Crown 8vo, cloth t Q() 

THE PRACTICAL ENGINEER'S HANDBOOK. 

Comprising a Treatise on Modern Engines and Boilers; Marine, Loco- 
motive, and Stationary. And containing a large collection of Rules and 
Practical Data relating to Recent Practice in Designing and Construct- 
ing all kinds of Engines, Boilers, and other Engineering Work. The 
whole constituting a comprehensive Key to the Board of Trade and 
other Examinations for Certificates of Competency in Modern Mechan- 
ical Engineering. By Walter S. Hutton, Civil and Mechanical En- 
gineer, Author of "The Works' Manager's Handbook for Engineers, " 
&c. With upwards of 420 Illustrations. Sixth Edition, Revised and 
Enlarged. Medium 8vo, nearly 560 pp., strongly bound $7.00 

t=P™ This Work is designed as a companion to the Author's "Works' 
Manager's Handbook." It possesses many new and original features, and 
contains, like its predecessor, a quantity of matter not originally intended for 
publication, but collected by the Author for his own use in the construction of a 
great variety of Modern Engineering Work. 

The information is given in a condensed and concise form, and is illus- 
trated by upwards cf 420 Engravings; and comprises a quantity of tabulated 
matter of great value to all engaged in designing, constructing, or estimating for 
Engines, Boilers, and other Engineering Work. 

TEXT-BOOK ON THE STEAM ENGINE. 

With a Supplement on Gas Engines and Part II. on Heat Engines 
By T. M. Goodeve, M.A., Barrister-at-Law, Professor of Mechanics at 
the Royal College of Science, London; Author of "The Principles of 
Mechanics," "The Elements of Mechanism," &c. Fourteenth Edition. 

Crown 8vo, cloth $2.00 

1 ' Professor Goodeve has given us a treatise on the steam engine which will 

bear comparison with anything written by Huxley or Maxwell, and we can 

award it no higher praise." — Engineer. 

A HANDBOOK ON THE STEAM ENGINE. 

With especial Reference to Small and Medium-sized Engines. For the 
Use of Engine Makers, Mechanical Draughtsmen, Engineering Students, 
and users of Steam Power. By Herman Haeder, C.E. Translated 
from the German, with additions and alterations, by H. H. P. Powles, 
A.M.LC.E., M.I.M.E. Third Edition, Revised. With nearly 1,100 

Illustrations. 12mo, cloth $3.00 

Summary of Contents: — Introduction. — Types of Steam Engines. — 
Details of Steam Engines. — Governors. — Valve Gears. — Condensers, 
Air-Pumps, and Feed-Pumps. — Examples of Engines of Continental 
Make, from Actual Practice. — Particulars of Engines by English 
Makers. — Compound Engines. — Indicator and Indicator Diagrams. — 
Calculations for Power and Steam Consumption. — Effect of Inertia 
on Reciprocating Parts of Engines. — Friction Brake Dynamometer — 
Sundry Details. — Boilers. — Index. 

" There can be no question as to its value. We cordially commend it 
to all concerned in the design and construction of the steam engine." — 
Mechanical World. 

THE PORTABLE ENGINE. 

A Practical Manual on its Construction and Management, for the use 
of Owners and Users of Steam Engines generally.- By William Dyson 

Wansbrough. 12mo, cloth $1.50 

"This is a work of value to those who use steam machinery. . . . Should 

be read by every one who has a steam engine, on a form or elsewhere ,"— ■» 

Mark <£<m« Expr&>*,*. 



6 0R05BY LOCKWOOD 6" SOX'S CATALOGUE. 
THE STEAM ENGINE. 

A Treatise on the Mathematical Theory of, with Rules and Examples 
for Practical Men. By T. Baker, C.E. 12mo, cloth (jO 

"Teems with scientific information with reference to the steam-engine." — 
Design and Work. 

THE STEAM ENGINE. 

For the use of Beginners. By Dr. Lardxer. 12mo, cloth. . . .(JO 

LOCOMOTIVE ENGINE DRIVING. 

A Practical Manual for Engineers in Charge of Locomotive Engines. 
By Michael Reynolds, M.S.E, TwellLh Edition. 12mo, cloth 
b °ards S2.00 

' ' We can confidently recommend the book, not only to the practical driver, 
but to even.- one who takes an interest in the performance of locomotive 
engines.'* — The Engineer. 

THE LOCOMOTIVE ENGINE. 

The Autobiography of an Old Locomotive Engine. By Robert 
Weathehburn, M.I.M.E. With Illustrations and Portraits of George 
and Robert Stephexsox. 12mo, cloth SI. 00 

THE LOCOMOTIVE ENGINE AND ITS DEVELOPMENT. 

A Popular Treatise on the Gradual Improvements made in Railway 
Engines between 1S03 and 1903. By Clement E. Strettox, C.E. 

Sixth Edition, Revised and Enlarged. 12mo, cloth S2.00 

"Students of railway history and all who are interested in the evolution 

of the modern locomotive will find much to attract and entertain in this 

volume." — The Times. 

THE MODEL LOCOMOTIVE ENGINEER, 

Fireman, and Engine-Boy. Comprising a Historical Notice of the 

Pioneer Locomotive Engines and their Inventors. By Michael Retx- 

olds. Second Edition, with Revised Appendix. 12mo, cloth. S2.00 

"We should be glad to see this book in the possession of every one in the 

kingdom who has ever laid, or is to lay, hands on a locomotive engine." — 

Iron. 

LOCOMOTIVE ENGINES. 

A Rudimentary Treatise on. By G. D. Dempset, C.E. With large 
•Additions treating of the Modern Locomotive, bv D. K. Clark, 

M.Inst. C.E. With Illustrations. 12mo, cloth S.120 

"A model of what an elementary technical book should be." — Academy. 

CONTINUOUS RAILWAY BRAKES. 

A Practical Treatise on the several Systems in Use in the United King- 
dom; their Construction and Performance. By M. Retxolds. 8vo, 
cloth 83.50 

ENGINE-DRIVING LIFE. 

Stirring Adventures and Incidents in the Lives of Locomotive Engine- 
Drivers. By Michael Retxolds. Third Edition. 12mo, cloth. ,60 

STATIONARY ENGINE DRIVING. 

A Practical Manual for Engineers in Charge of Stationary Engines. By 
Michael Retxolds, M.S.E. Seventh Edition. 12mo, cloth boards. 

S2.00 

THE CARE AND MANAGEMENT OF STATIONARY 

ENGINES. 

A Practical Handbook for Men-in-charge. By C. Hurst. 12mo. ,50 



MECHANICAL ENGINEERING, &c. 7 

THE ENGINEMAN'S POCKET COMPANION 

and Practical Educator for Enginemen, Boiler Attendants, and Me- 
chanics. By Michael Reynolds. With 45 Illustrations and numer- 
ous Diagrams. Fifth Edition. Royal 18mo, strongly bound for 

Pocket wear $1.50 

"A most meritorious work, giving in a succinct and practical form all the 

information an engine-minder, desirous of mastering the scientific principles 

of his daily calling, would require." — The Miller., 

THE SAFE USE OF STEAM. 

Containing Rules for Unprofessional Steam Users. By an Engineer. 

Eighth Edition. Sewed .25 

"If steam-users would but learn this little book by heart, boiler explo- 
sions would become sensations by their rarity." — English Mechanic. 

STEAM AND MACHINERY MANAGEMENT. 

A Guide to the Arrangement and Economical Management of Machin- 
ery, with Hints on Construction and Selection. By M. Powis Bale, 
M.Inst.M.E. 12mo, cloth $1.00 

GAS AND OIL ENGINE MANAGEMENT. 

A Practical Guide for Users and Attendants, being Notes on Selection, 
Construction, and Management. By M. Powis Bale, M.Inst.C.E., 
M.I.Mech.E. Author of "Woodworking Machinery," &c. 12mo, 
cloth $1.50 

ON GAS ENGINES. 

With Appendix describing a Recent Engine with Tube Igniter. By 
T. M. Goodeve, M.A. 12mo, cloth $1.00 

THE ENGINEER'S YEAR=BOOK FOR 1906. 

Comprising Formulae, Rules, Tables, Data, and Memoranda in Civil, 
Mechanical, Electrical, Marine, and Mine Engineering. By H. R. 
Kempe, M.Inst.C.E., Principal Staff Engineer, Engineer-in-Chief's 
Office, General Post Office, London; Author of "A Handbook of Elec- 
trical Testing," "The Electrical Engineer's Pocket-Book," &c. With 
1,000 Illustrations, specially Engraved for the Work. 12mo, 950 pp., 
leather $3.00 

THE MECHANICAL ENGINEER'S POCKET=BOOK. 

Comprising Tables, Formulae, Rules, and Data: a Handy Book of Ref- 
erence for Daily Use in Engineering Practice. By D. Kinnear Clark, 
M.Inst.C.E. , Fifth Edition, thoroughly Revised and Enlarged. By H. H. 
P. Powles, A.M.Inst.C.E., M.I.M.E. Small 8vo, 700 pp., leather. $3, QQ 

Summary of Contents: — Mathematical Tables. — Measurement of 
Surfaces and Solids. — English Weights and Measures. — French 
Metric Weights and Measures. — Foreign Weights and Measures. — 
Moneys. — Specific Gravity, Weight, and Volume. — Manufactured 
Metals. — Steel Pipes. — Bolts and Nuts. — Sundry Articles in Wrought 
and Cast Iron, Copper, Brass, Lead, Tin, Zinc. — Strength of Mater- 
ials. — Strength of Timber. — Strength of Cast Iron. — Strength of 
Wrought Iron. — Strength of Steel. — Tensile Strength of Copper, 
Lead, &c. — Resistance of Stones and other Building Materials. — 
Riveted Joints in Boiler Plates. — Boiler Shells. — Wire Ropes and 
Hemp Ropes — Chains and Chain Cables. — Framing. — Hardness of 
Metals, Alloys, and Stones. — Labour of Animals. — Mechanical Prin- 
ciples. — Gravity and Fall of Bodies. — Accelerating and Retarding 
Forces. — Mill Gearing, Shafting, &c. — Transmission of Motive Power. 
— Heat. — Combustion. — Fuels. — Warming, Ventilation, Cooking 
Stoves. — Steam. — Steam Engines and Boilers. — Railways. — Tram- 
ways. — Steam Ships. — Pumping Steam Engines and Pumps. — Coal Gas, 
Gas Engines, &c. — Air in Motion. — Compressed Air. — Hot- Air Engines. 
— Water Power. — Speed of Cutting Tools. — Colours. — Electrical 
Engineering. 



8 CROSBY LOCKWOOD 6 s SON'S CATALOGUE. 
PRACTICAL MECHANICS' WORKSHOP COMPANION. 

Comprising a great Variety of the most useful Rules and Formula? in 
Mechanical Science, with numerous Tables of Practical Data and Cal- 
culated Results for Facilitating Mechanical Operations. By William 
Templeton, Author of "The Engineer's Practical Assistant," &c, &c. 
Eighteenth Edition, Revised, Modernised, and considerably Enlarged, 
by W. S. Hutton, C.E., Author of "The Works' Manager's Hand- 
book," &c. Fcap. 8vo, nearly 500 pp., with 8 Plates and upwards of 
250 Diagrams, leather $2.50 

ENGINEER'S AND MILLWRIGHT'S ASSISTANT. 

A Collection of Useful Tables, Rules, and Data. By William Temple- 
ton. Eighth Edition, with Additions. 18mo, cloth SI. 00 

TABLES AND MEMORANDA FOR ENGINEERS, 

MECHANICS, ARCHITECTS, BUILDERS, &c. 

Selected and Arranged by Francis Smith. Seventh Edition, Revised, 
including Electrical Tables, Formula, and Memoranda. Waist- 
coat-pocket size, limp leather .60 

THE MECHANICAL ENGINEER'S COMPANION. 

Of Areas, Circumferences, Decimal Equivalents, in inches and feet, mil- 
limetres, squares, cubes, roots, &c; Strength of Bolts, Weight of Iron, 
&c; Weights, Measures, and other Data. Also Practical Rules for 
Eneane Proportions. By R. Edwards, M.Inst.C.E. Fcap. 8vo, cloth. 

SI. 00 
MECHANICAL ENGINEERING TERMS. 

(Lockwood's Dictionary of). Embracing those current in the Drawing 
Office, Pattern Shop, Foundry, Fitting, Turning, Smiths', and Boiler 
Shops, &c. Comprising upwards of 6,000 Definitions. Edited by J. 
G Horner, A.M.I.M.E. Third Edition, Revised, with Additions. 
12mo, cloth S3.00 

"Just the sort of handy dictionary required by the various trades engaged 
in mechanical engineering. The practical engineering pupil will find the 
book of great value in his studies, and every foreman engineer and mechanic 
should have a copy." 

POCKET GLOSSARY OF TECHNICAL TERMS. 

English-French, French-English ; with Tables suitable for the Archi- 
tectural, Engineering, Manufacturing, and Nautical Professions. By 
John James Fletcher. Fourth Edition, 200 pp. Waistcoat -pocket 
size, limp leather .60 

IRON AND STEEL. 

A Work for the Forge Foundry, Factory, and Office. Containing ready, 
useful, and trustworthy Information for Ironmasters and their Stock- 
takers'; Managers of Bar, Rail, Plate, and Sheet Rolling Mills; Iron and 
Metal Founders; Iron, Ship, and Bridge Builders; Mechanical, Mining, 
and Consulting Engineers; Architects, Contractors, Builders, &c. By 
Charles Hoare, Author of "The Slide Rule," &c. Ninth Edition. 
32mo, leather S2.50 

WORKMAN'S MANUAL OF ENGINEERING DRAWING. 

By John Maxton, Instructor in Engineering Drawing, Royal Naval 
College, Greenwich. Eighth Edition. 300 Plates and Diagrams. 

12mo, cloth S1.40 

"A copy of it should be kept for reference in every drawing office." — En- 
gineering. 

PATTERN MAKING. 

Embracing the Main Types of Engineering Construction, and including 
Gearing, Engine Work, Sheaves and Pulleys, Pipes and Columns, Screws, 
Machine Parts, Pumps and Cocks, the Moulding of Patterns in Loam 
and Greensand, Weight of Castings, &c. By J. G. Horner, A.M.I.M.E. 
Third Edition, Enlarged. With 486 Illustrations. 12mo, cloth. S3.00 



MECHANICAL ENGINEERING, &c. g 

SMITHY AND FORGE. 

Including the Farrier's Art and Coach Smithing. By W. J. E. Crane. 

12mo, cloth $1.00 

"The first modern English book on the subject. Great pains have been 
bestowed by the author upon the book; shoeing-smiths will find it both 
useful and interesting." 

TOOTHED GEARING. 

A Practical Handbook for Offices and Workshops. By J. Horner, 
A.M.I.M.E. Second Edition, with a new Chapter on Recent Practice. 
With 184 Illustrations. 12mo, cloth $2.25 

MODERN WORKSHOP PRACTICE, 

As applied to Marine, Land, and Locomotive Engines, Floating Docks, 
Dredging Machines, Bridges, Shipbuilding, &c. By J. G. Winton. 
Fourth Edition, Illustrated. 12mo, cloth $1.40 

DETAILS OF MACHINERY. 

Comprising Instructions for the Execution of various Works in Iron in 
the Fitting Shop, Foundry, and Boiler Yard. By Francis Campin, 
C.E. 12mo, cloth $1.20 

ENGINEERING ESTIMATES, COSTS, AND ACCOUNTS. 

A Guide to Commercial Engineering. With numerous examples of Es- 
timates and Costs of Millwright Work, Miscellaneous Productions, 
Steam Engines and Steam Boilers; and a Section on the Preparation 
of Costs Accounts. By A General Manager. Second Edition. 8vo, 
cloth $4.50 

MECHANICAL ENGINEERING. 

Comprising Metallurgy, Moulding, Casting, Forging, Tools, Workshop 
Machinery, Mechanical Manipulation, Manufacture of the Steam En- 
gine, &c. By Francis Campin, C.E. Third Edition. 12mo, cloth 

$1.00 
LATHE=WORK. 

A Practical Treatise on the Tools. Appliances, and Processes employed in 
the Art of Turning. By Paul N. Hasltjck. Eighth Edition. 12mo, 

cloth $2.00 

"Written by a man who knows not only how work ought to be done, but 

who also knows how to do it, and how to convey his knowledge to others." — 

Engineering. 

SCREW=TH READS, 

And Methods of Producing Them. With numerous Tables and com- 
plete Directions for using Screw-cutting Lathes. By Paul N. Hasluck. 
Author of "Lathe-work," &c. Sixth Edition. Waistcoat-pocket size. 

.60 

"Full of useful information, hints and practical criticism. Taps, dies, 
and screwing tools generally are illustrated and their action described." 

CONDENSED MECHANICS. 

A Selection of Formula?, Rules, Tables, and Data for the Use of Engi- 
neering Students, &c. By W. G. C. Hughes, A.M.I.C.E. 12mo, cloth. 

$1.00 
MECHANICS OF AIR MACHINERY. 

By Dr. J. Weisbach and Prof. G. Herrmann. Authorized Translation 
with an Appendix on American Practice by A. Trowbridge, Ph.B., 
Adjunct Professor of Mechanical Engineering, Columbia University. 
Royal 8vo, cloth. Net $3.75 



io CROSBY L0CKW00D 6- SON'S CATALOGUE. 
PRACTICAL MECHANISM. 

And Machine Tools. By T. Baker, C.E. With Remarks on Tools and 
Machinery by J. Nasmyth, C.E. 12mo, cloth $1.00 

MECHANICS. 

Being a concise Exposition of the General Principles of Mechanical 
Science and their Applications. By C. Tomlinson, F.R.S. 12mo, 
cloth .60 

FUELS: SOLID, LIQUID, AND GASEOUS. 

Their Analysis and Valuation. For the use of Chemists and Engineers. 
By H. J. Phillips, F.C.S., formerly Analytical and Consulting Chemist 
to the Great Eastern Railway. Fourth Edition. 12mo, cloth. . ,gQ 

"Ought to have its place in the laboratory of every metallurgical estab- 
lishment and wherever fuel is used on a large scale." — Chemical News. 

FUEL, ITS COMBUSTION AND ECONOMY. 

Consisting of an Abridgment of "A Treatise on the Combustion of Coal 
and the Prevention of Smoke." By C. W. Williams, A. Inst. C.E. 
With extensive Additions by D. Kinnear Clark, M.Inst. C.E. 
Fourth Edition. 12mo, cloth $1.50 

"Students should buy the book and read it, as one of the most complete 
and satisfactory treatises on the combustion and economy of fuel to be 
had . ' ' — Engineer. 

STEAM AND THE STEAM ENGINE, 

Stationary and Portable. Being an Extension of the Treatise on the 
Steam Engine of Mr. J. Sewell. By D. K. Clark, C.E. Fourth Edi- 
tion. 12mo, cloth $1.40 

"Every essential part of the subject is treated of competently, and in a 
popular style." 

PUMPS AND PUMPING. 

A Handbook for Pump Users. Being Notes on Selection, Construction, 
and Management. By M. Powis Bale, M.Inst. C.E. , M.I.Mech.E. 

Fourth Edition. 12mo», cloth $1.50 

" Thoroughly practical and clearly written." 

REFRIGERATION, COLD STORAGE, & ICE-MAKING. 

A Practical Treatise on the Art and Science of Refrigeration. By. A. 
J. Wallis-Tayler, A.M. Inst. C.E. , Author of "Refrigerating and Ice- 
Making Machinerv." 600 pp., with 360 Illustrations. Medium 8vo, 

cloth $4.50 

Contents: — Chapter I. Introduction. — II. The Theory and Pracs 
tice op Refrigeration. — III. The Liquefaction Process. — IV. The 
Vacuum Process. — V. The Compression Process or System. — VI. The 
Compression Process (Continued). — VII. The Compression Process (Con- 
tinued). — VIII. Condensers and Water-Cooling and Saving Apparatus. 
— IX. The Absorption and Binary Absorption Process or System. — 
X. The Cold-Air System. — XI. Cocks, Valves and Pipe-Joints and 
Unions. — XII. Refrigeration and Cold Storage. — XIII. Refrigera- 
tion and Cold Storage (Continued). — XIV. Refrigeration and Cold 
Storage (Continued). — XV. Refrigeration and Cold Storage (Con- 
tinued). — XVI. Marine Refrigeration. — XVII. Manufacturing, In- 
dustrial and Constructional Applications. — XVIII. Ice-Making. — 

XIX. The Management and Testing of Refrigerating Machinery. — 

XX. Cost of Working. — XXI. The Production of Very Low Temper- 
atures. — XXII. Useful Tables and Memoranda. — Appendix. — Bibli' 

OGRAPHY OF REFRIGERATION. 



CIVIL ENGINEERING, SURVEYING, &c. n 

THE POCKET BOOK OF REFRIGERATION AND ICE- 
MAKING. 

By A. J. Wallis-Tayler, A.M. Inst. C.E. Author of "Refrigerating 
and Ice-making Machinery," &c. Third Edition, Enlarged. 12mo, 
cloth $1.50 

REFRIGERATING & ICE=MAKING MACHINERY. 

A Descriptive Treatise for the Use of Persons Employing Refrigerating 
and Ice-making Installations, and others. By A. J. Wallis-Tayler, 

A.M.Inst.C.E. Third Edition, Enlarged. 12mo, cloth $3.00 

"May be recommended as a useful description of the machinery, the proc- 
esses, and of the acts, figures, and tabulated physics of refrigerating." — En- 
gineer. 

MOTOR VEHICLES FOR BUSINESS PURPOSES. 

A Practical Handbook for those interested in the Transport of Passen- 
gers and Goods. By A. J. Wallis-Tayler, A.M.Inst.C.E. With 134 
Illustrations. Demy 8vo, cloth [Just -published. ,] $3.50 

MOTOR CARS OR POWER=CARRIAGES FOR COMMON 

ROADS. 

By A. J. Wallis-Tayler, A.M.Inst.C.E. 212 pp., with 76 Illustrations. 
12mo, cloth $2.00 

AERIAL NAVIGATION. 

A Practical Handbook on the Construction of Dirigible Balloons, Aero- 
stats, Aeroplanes, and Aeromotors. By Frederick Walker, C.E., 
Associate Member of the Aeronautic Institute. With 104 Illustrations. 
Large 12mo, cloth $3.00 

STONE=WORKING MACHINERY. 

A Manual dealing with the Rapid and Economical Conversion of Stone. 
With Hints on the Arrangement and Management of Stone Works. By 
M. Powis Bale, M.Inst.C.E., M.I.Mech.E. Second Edition, enlarged. 

12mo, cloth $3.50 

"The book should be in the hands of every mason or student of stone- 
work." 

"A handbook for all who manipulate stone for building or ornamental 
purposes." 

FIRES, FIRE=ENGINES, AND FIRE BRIGADES. 

With a History of Fire-Engines, their Construction, Use, and Manage- 
ment; Foreign Fire Systems; Hints on Fire-Brigades, &c. By C. F. 
T. Young, C.E. 8vo, cloth $8.00 

CRANES. 

The Construction of, and other Machinery for Raising Heavy Bodies 
"for the Erection of Buildings, &c. By J. Glynn, F.R.S. 12mo, cloth. 

.60 



CIVIL ENGINEERING, SURVEYING, ETC. 
PIONEER IRRIGATION. 

A Manual of Information for Farmers in the Colonies. By E. 0. Maw- 
son, M.Inst.C.E., Executive Engineer, Public Works Department, 
Bombay. With Additional Chapters on Light Railways by E. R. 
Calthrop, M.Inst.C.E., M.I.M.E. Illustrated by numerous Plates 

and Diagrams. Demy 8vo, cloth $4.00 

Summary of Contents: — Value of Irrigation, and Sources op Water 
Supply. — Dams and Weirs. — Canals. — Underground Water. — Meth- 
ods of Irrigation. — Sewage Irrigation. — Imperial Automatic Sluice 
Gates. — The Cultivation of Irrigated Crops, Vegetables, and Fruit 
Trees. — Light Railways for Heavy Traffic. — Useful Memoranda and 
Data. 



12 CROSBY LOCKWOOD & SOX'S CATALOGUE. 
THE RECLAMATION OF LAND FROM TIDAL WATERS. 

A Handbook for Engineers, Landed Proprietors, and others interested 
in Works of Reclamation. By A. Beazely, M.Inst.C.E. 8vo, cloth. 

S4.00 
"The book shows in a concise way what has to be done in reclaiming land 
from the sea, and the best way of doing it. Contains a great deal of prac- 
tical and useful information which cannot fail to be of service to engineers 
entrusted with the enclosure of salt marshes, and to landowners intending 
to reclaim land from the sea." — The Engineer. 

THE WATER SUPPLY OF TOWNS AND THE CON- 
STRUCTION OF WATER-WORKS. 

A Practical Treatise for the Use of Engineers and Students of Engineer- 
ing. By W. K. Botox, A. M.Inst.C.E., Consulting Engineer to the 
Tokyo Water-works. Second Edition, Revised and Extended. With 
numerous Plates and Illustrations. Super-royal Svo, buckram. S9.00 

I. INTRODUCTORY. II. DIFFERENT QUALITIES OF WATER. III. QUAN- 
TITY of Water to be Provided. — IV. On Ascertaining whether a Pro- 
posed Source of Supply is Sufficient. — V. On Estimating the Storage 
Capacity Required to be Provided. — VI. Classification of Water- 
works. — VII. Impounding Reservoirs. — VIII. Earthwork Dams. — IX. 
Masonry Dams. — X. The Purification of Water. — XI. Settling Res- 
ervoirs. — XII. Sand Filtration. — XIII. Purification of Water by 
Action of Iron, Softening of Water by Action of Lime, Xatural 
Filtration. — XIV. Service or Clean Water Reservoirs — Water 
Towers — Stand Pipes. — XV. The Connection of Settling Reservoirs, 
Filter Beds and Service Reservoirs. — XVI. Pumping Machinery. — 

XVII. Flow of Water in Conduits — Pipes and Open Channels. — 

XVIII. Distribution Systems. — XIX. Special Provisions for the Ex- 
tinction of Fire. — XX. Pipes for Water-works. — XXI. Prevention 
of Waste of Water. — XXII. Various Appliances used in Connection 
with Water-works. 

Appendix I. By Prof. JOHN MILXE. F.R.S.— Considerations Con- 
cerning the Probable Effects of Earthquakes on Water-works, and 
the Special Precautions to be Taken in Earthquake Countries. 

Appendix II. By JOHN DE RIJKE, C.E.— On Sand Dunes and Dune 
Sand as a Source of Water Supply. 

THE WATER SUPPLY OF CITIES AND TOWNS. 

By William Humber. A. M.Inst.C.E., and M.Inst.M.E., Author of 
"Cast and Wrought Iron Bridge Construction," &c, &c. Illustrated 
with 50 Double Plates. 1 Single Plate. Coloured Frontispiece, and up- 
wards of 250 Woodcuts, and containing 400 pp. of Text. Imp. 4to, 

elegantly and substantially half-bound in morocco S45.00 

List of Contents: — I. Historical Sketch of some of the means that 

HAVE BEEN ADOPTED FOR THE SUPPLY OF WATER TO CrTIES AND TOWNS. 

IL Water and the Foreign Matter usually associated with it. — III. 
Rainfall and Evaporation. — IV. Springs and the Water-bearing 
Formations of Various Districts. — V. Measurement and Estimation 
of the Flow of Water. — VI. On the Selection of the Source of Sup- 
ply. — VII. Wells. — VIII. Reservoirs. — IX. The Purification of 
Water. — X Pumps. — XI. Pumping Machinery. — XII. Conduits. — 
XIII. Distribution of Water. — XIV. Meters. Service Pipes, and 
House Fittings. — XV. The Law and Economy of Water-works. — XVI. 
Constant and Intermittent Supply. — XVII. Description of Plates. — 
Appendices, giving Tables of Rates of Supply. Velocities, &c, &c, 
together with specifications of several works illustrated. among 
which will be found: aberdeen. bldeford, canterbury, dundee, 
Halifax, Lambeth, Rotherham, Dublin, and others. 

RURAL WATER SUPPLY. 

A Practical Handbook on the Simply of Water and Construction of 
Water-works for small Count rv Disfricts. Bv Allan Greenwell, 
A .M.Inst.C.E., and W. T. Curry. A.M Inst.C.E., F.G.S. With Illus- 
te-»tiene, Seeesd Edition. Revised. 12mo. clotfe. ........... $2.QO 



CIVIL ENGINEERING, SURVEYING, &c. 13 

WATER ENGINEERING. 

A Practical Treatise on the Measurement, Storage, Conveyance, and 
Utilization of Water for the Supply of Towns, for Mill Power, and for 
other Purposes. By Charles Slagg, A.M.Inst.C.E. Second Edition. 
12mo, cloth $3.00 

WATER WORKS, FOR THE SUPPLY OF CITIES AND 

TOWNS. 

With a Description of the Principal Geological Formations of England 
as influencing Supplies of Water. By Samuel Hughes. 12mo, cloth 

$1.60 
POWER OF WATER. 

As applied to drive Flour Mills, and to give motion to Turbines, and 
other Hydrostatic Engines. By Joseph Glynn, F.R.S., &c. New 
Edition. Illustrated. 12mo, cloth gq 

WELLS AND WELL=SINKING. 

By J. G. Swindell, A.R.I.B.A., and G. R. Burnell, C.E. Revised 
Edition. 12mo, cloth ^gQ 

"Solid practical information, written in a concise and lucid style. The 
work can be recommended." 

HYDRAULIC POWER ENGINEERING. 

A Practical Manual on the Concentration and Transmission of Power 
by Hydraulic Machinery. By G. Croydon Marks, A.M.Inst.C.E. 
Second Edition, Enlarged, with about 240 Illustrations. 8vo, cloth. 

[Just Published. $3.50 
Summary of Contents: — Principles of Hydraulics. — The Flow of 
Water. — Hydraulic Pressures. — Material. — Test Load. — Packings 
for Sliding Surfaces. — Pipe Joints. — Controlling Valves. — Platform 
Lifts. — Workshop and Foundry Cranes. — Warehouse and Dock 
Cranes. — Hydraulic Accumulators. — Presses for Baling and other 
Purposes. — Sheet Metal Working and Forging Machinery. — Hy- 
draulic Riveters. — Hand and Power Pumps. — Steam Pumps. — Tur- 
bines. — Impulse Turbines. — Reaction Turbines. — Design of Tur- 
bines in Detail. — Water Wheels. — Hydraulic Engines. — Recent 
Achievements. — Pressure of Water. — Action of Pumps, &c. 

HYDRAULIC MANUAL. 

Consisting of Working Tables and Explanatory Text. Intended as a 
Guide in Hydraulic Calculations and Field Operations. By Lowis 
D'A. Jackson, Author of "Aid to Survey Practice," "Modern Metrol- 
ogy," &c. Fourth Edition, Enlarged. 8vo, cloth $6.00 

"The author has constructed a manual which may be accepted as a trust- 
worthy guide to this branch of the engineer's profession." — Engineering. 

HYDRAULIC TABLES, CO=EFFICIENTS, & FORMUL/E. 

For Finding the Discharge of Water from Orifices, Notches, Weirs, 
Pipes, and Rivers. With New Formula?, Tables, and General Informa- 
tion on Rain-fall, Catchment-Basins, Drainage, Sewerage, Water Sup- 
ply for Towns and Mill Power. By John Neville, C.E. , M.R.I.A- 
Third Edition, revised, with additions. Numerous Illustrations. 
12mo, cloth. $5.00 

"It is, of all English books on the subject, the one nearest to complete- 
ness." 

MASONRY DAMS FROM INCEPTION TO COMPLETION. 

Including numerous Formula?, Forms of Specifications and Tender, 
Pocket Diagram of Forces, &c. For the use of Civil and Mining En- 
gineers. By C. F. Courtney, M.Inst.C.E. 8vo, cloth $3.50 

"Contains a good deal of valuable data. Many useful suggestions will be 

found in the remarks on site and position, location of darn, foundations 

and constuct'on." — Building News. 



14 CROSBY LOCKWOOD & SON'S CATALOGUE. 
RIVER BARS. 

The Causes of their Formation, and their Treatment by "Induced Tidal 
Scour"; with a Description of the Successful Reduction by this Method 
of the Bar at Dublin. By I. J. Mann, Assist. Eng. to the Dublin Port 

and Docks Board. Royal 8vo, cloth $3.00 

"We recommend all interested in harbour works — and, indeed, (hose con- 
cerned in the improvements of rivers generally — to read Mr. Mann's inter- 
esting work." — Engineer. 

DRAINAGE OF LANDS, TOWNS, AND BUILDINGS. 

By G. D. Dempsey, C.E. Revised, with large Additions on Recent 
Practice in Drainage Engineering by D. Kinnear Clark, M.fnst.C.E. 
Fourth Edition. 12mo, cloth $1.80 

i 

SURVEYING AS PRACTISED BY CIVIL ENGINEERS 

AND SURVEYORS. 

Including the Setting-out of Works for Construction and Surveys 
Abroad, with many Examples taken from Actual Practice. A Hand- 
book for use in the Field and the Office, intended also as a Text -book 
for Students. By John Whitelaw, Jun., A. M.Inst. C.E. , Author of 
"Points and Crossings." With about 260 Illustrations. Demy 8vo, 
cloth $4.00 

PRACTICAL SURVEYING. 

A Text-book for Students preparing for Examination or for Sur-vey- 
work in the Colonies. By C.eorge W. Usiel, A.M.Inst.C.E. Eighth 
Edition, thoroughly Revised and Enlarged, ry Aiex Beazeley, 
M.Inst.C.E. With 4 Lithographic Plates and 360 Illustrations. 12mo, 
cloth $3.00 

SURVEYING WITH THE TACHEOMETER. 

A practical Manual for the use of Civil and Military Engineers and Sur- 
veyors, including two series of Tables specially computed for the Re- 
duction of Readings in Sexagesimal and in Centesimal Degrees. By 
Neie Kennedy, M.Inst.C.E. With Diagrams and Plates. Second 

Edition. 8vo, cloth -$4.00 

"The work is very clearly written, and should remove all difficulties in the 
way of any surveyor desirous of making use of this useful and rapid instru- 
ment . ' ' — Nature. 

LAND AND ENGINEERING SURVEYING. 

For Students and Practical Use. Py T. Baker, C.E. Twentieth Edi- 
tion, by F. E. Dixon, A.M.Inst.C.E. With Plates and Diagrams. 
12mo, cloth 80 

AID TO SURVEY PRACTICE. 

For Reference in Surveying, Levelling, and Setting-out; and in Route 
Surveys of Travellers by Land and Sea. With Tables, Illustrations, 
and Records. By L. D'A. Jackson, A.MJnst.C.E. Second Edition. 
8vo, cloth $5.00 

LAND AND MARINE SURVEYING. 

In Reference to the Preparation of Plans for Roads and Railways; 
Canals, Rivers, Towns' Water Supplies; Docks and Harbours. With 
Description and Use of Surveying Instruments. By W. Davis Haskoli., 
C.E. Second Edition , Revised with Additions. Crown 8vo, cloth . 

$3.50 



CIVIL ENGINEERING, SURVEYING, &c. 15 

ENGINEER'S & MINING SURVEYOR'S FIELD B( £>K. 

Consisting of a Series of Tables, with Rules, Explanations of Systems 
and use of Theodolite for Traverse Surveying- and plotting the work 
with minute accuracy by means of Straight Edge and Set Square only; 
Levelling with the Theodolite, Setting-out Curves with and without the 
Theodolite, Earthwork Tables, &c. By W. Davis Haskoll, C.E. With 
numerous Woodcuts. Fifth Edition, Enlarged. J2mo, cloth. $4.50 
"The book is very handy; the separate tables of sines and tangents to 

every minute will make it useful for many other purposes, the genuine 

traverse tables existing all the same." 

AN OUTLINE OF THE METHOD OF CONDUCTING 

A TRIGONOMETRICAL SURVEY. 

For the Formation of Geographical and Topographical Maps and Plans, 
Military Reconnaissance, LEVELLING, &c, with Useful Problems, 
Formulae, and Tables. By Lieut. -General Frome, R.E. ]<ourth Edi- 
tion, Revised and partly Re-written by Major-General Sir Charles 
Warren, G.C.M.G., R.E. With 19 Plates and 115 Woodcuts. 8vo, 

doth $6.00 

PRINCIPLES AND PRACTICE OF LEVELLING. 

Showing its Application to Purposes of Railway and Civil Engineering 
in the Construction of Roads; with Mr. Telford's Rules for the same. 
By Frederick W. Simms, M.Inst. C.E. Eighth Edition, with Law's 
Practical Examples for Setting-out Railway Curves, and Trautwine's 
Field Practice of Laying out Circular Curves. With 7 Plates and nu- 
merous Woodcuts. 8vc $2.50 

"The text-book on levelling in most of our engineering schools and col- 
leges."- -Engineer. 

"The publishers have rendered a substantial service to the profession, 
especially to the younger members, by bringing out the present edition of 
Mr. Simm's useful work." — Engineering. 

TABLES OF TANGENTIAL ANGLES AND MULTIPLES. 

For Setting-out Curves from 5 to 200 Radius. By A. BEAZELEY.M.Tnst. 
C.E. 7th Edition. Revised. With an Appendix on the use of the 
Tables for Measuring up Curves. Printed on 50 Cards, and sold in a 

cloth box, waistcoat-pocket size $1.50 

"Each table is printed on a small card, which, placed on the theodolite, 

leaves the hands free to manipulate the instrument — no small advantage 

as regards the rapidity of work." 

"Very handy; aman may know that all his day's work must fall on two 

of these cards, which he puts into his own card-case, and leaves the rest 

behind." 

PIONEER ENGINEERING. 

A treatise on the Engineering Operations connected with the Settle- 
ment of Waste Lands in New Countries. By E. Dobson, M.Inst. C.E. 

Second Edition. 12mo, cloth $1.80 

"Mr Dobson is familiar with the difficulties which have to be overcome 

in this class of work, and much of his advice will be valuable to young 

engineers proceeding to our colonies." — Engineering. 

TUNNELLING. 

A Practical Treatise. By Charles Prelini, C.E With additions by 
Charhos S. Hill, C.E. With 150 Diagrams and Illustrations. Royal 
8vo, cloth $3.00 

PRACTICAL TUNNELLING. 

Explaining in detail Setting-out the Works, Shaft-sinking, and Heading- 
driving, Ranging the Lines and Levelling underground, Sub-Excavat- 
ing, Timbering and the Construction of the Brickwork of Tunnels. By 
F. W. Simm?., M.Inst. C.E. Fourth Edition. Revised and Further Ex- 
tended, including the most recent (1895) Examples of Sub-aqueous and 
other Tunnels, by D Kinnear Clark, M Inst C.E With 34 Folding 
Plates Imperial 8vo, cloth $9.00 



16 CROSBY LOCKWOOD ^ SOX'S CATALOGUE. 
EARTH AND ROCK EXCAVATION. 

A Practical Treatise, by Charles Prelini, C.E. 365 pp.. with Tables, 
manv Diagrams and Engravings. Roval Svo, cloth. 

[Just Published. §3.00 

CONSTRUCTION OF ROADS AND STREETS. 

By H. Law. C.E.. and D. K. Clark, C.E. Sixth Edition, revised, with 
Additional Chapters bv A. J. Wallis-Taylee, A.M. Inst. C.E. 12mo, 
cloth S2.50 

TRAMWAYS: THEIR CONSTRUCTION AND WORKING. 

Embracing a Comprehensive History of the System ; with an exhaustive 
Analysis of the Various Modes of Traction, including Horse Power. 
Steam. Cable Traction, Electric Traction. &c; a Description of the 
Varieties of Rolling Stock; and ample Details of Cost and Working 
Expenses. New Edition, Thoroughly Revised, and Including the 
Progress recently made in Tramway Construction. &e., <xc. By D. 
Ktnnear Clark, M.Inst. C.E. With 400 Illustrations. Svo, 780 pp., 
buckram 87.50 

HANDY GENERAL EARTH=WORK TABLES. 

Giving the Contents in Cubic Yards of Centre and Slopes of Cuttings 
and Embankments from 3 inches to 80 feet in Depth or Height, for use 
with either 66 feet Chain or 100 feet Chain. By J. H. Watson Beck, 
M.Inst. C.E. On a sheet mounted in cloth case SI. 50 

EARTHWORK TABLES. 

Showing the Contents in Cubic Yards of Embankments. Cuttings. &c, 
of Heights or Depths up to an average of 80 f eet. Bv Joseph Broad- 
bent, C.E., and Francis Campin. C.E. 12mo, cloth S2.00 

''The way in which accuracy is attained, by a simple division of each cross 

section into three elements, two m which are constant and one variable, is 

ingenious." 

A MANUAL ON EARTHWORK. 

Bv Alex. J. Graham, C.E. With numerous Diagrams. Second Edi- 
tion. 18mo, cloth SI. 00 

THE CONSTRUCTION OF LARGE TUNNEL SHAFTS. 

A Practical and Theoretical Essay. By J. H. Watson Beck, M.Inst. 

C.E., Resident Engineer. L. and N. W. R. With Folding Plates, Svo. 

cloth S4.80 

"Many of the methods given are of extreme practical value to the mason, 
and the "observations on the form of arch, the rules for ordering the stone, 
and the construction of the templates, will be found of considerable use," 

ESSAY ON OBLIQUE BRIDGES 

(Practical and Theoretical.) With 13 large Plates. By the late George 
Watson Beck. M. Inst. C.E Fourth Edition, revised by his Son, J H. 
Watson Beck. M.Inst C.E ; and with the addition of Description to 
Diagrams for Facilitating the Construction of Oblique Bridges, by ^ . 
H. Barlow, M.Inst.C.E Royal 8vo. cloth S4.S0 

CAST & WROUGHT IRON BRIDGE CONSTRUCTION 

(A Complete and Practical Treati^ on\ including Iron Foundations 
In Three Parts. — Theoretical, Practical, and Descriptive By Wil- 
liam Hember, A.M.Inst C.E . and M Inst.M.E. Third Edition, revised 
and much improved, with 115 Double Plates (20 of which now first 
appear in this edition ), and numerous Additions to the Text In 2 vols., 
imp. 4to. half-bound in morocco S50.00 

IRON BRIDGES OF MODERATE SPAN: 

Their Construction and Erection By H. W. Pendred With 40 Il- 
lustrations. 12mo, cloth ,80 



CIVIL ENGINEERING, SURVEYING, &c. 17 

IRON AND STEEL BRIDGES AND VIADUCTS. 

A Practical Treatise upon their Construction. For the use of Engi- 
neers, Draughtsmen, and Students. By Francis Campin, C.E. 12mo, 
cloth $1.40 

TUBULAR AND OTHER IRON GIRDER BRIDGES, 

Describing the Britannia and Conway Tubular Bridges. With a 
Sketch of Iron Bridges, &c. By G. D. Dempsey, C.E. 12mo, cloth, 

.80 

GRAPHIC AND ANALYTIC STATICS. 

In Their Practical Application to the Treatment of Stresses in Roofs, 
Solid Girders, Lattice, Bowstring, and Suspension Bridges, Braced 
Iron Arches and Piers, and other Frameworks. By R. Hudson 
Graham, C.E. Containing Diagrams and Plates to Scale. With num- 
erous Examples, many taken from existing Structures. Specially 
arranged for Class-work in Colleges and Universities. Second Edition, 
Revised and Enlarged. 8vo, cloth $6.00 

WEIGHTS OF WROUGHT IRON & STEEL GIRDERS. 

A Graphic Table for Facilitating the Computation of the Weights of 
Wrought Iron and Steel Girders, &c, for Parliamentary and other 
Estimates. By J. H. Watson Buck, M.Inst.C.E. On a sheet. $1.00 

GEOMETRY FOR TECHNICAL STUDENTS. 

An Introduction to Pure and Applied Geometry and the Mensuration 
of Surfaces and Solids, including Problems in Plane Geometry useful in 
Drawing. By E. H. Sprague, A.M.Inst.C.E. 12mo, cloth. . . .50 

PRACTICAL GEOMETRY. 

For the Architect, Engineer, and Mechanic. Giving Rules for the Delin- 
eation and Application of various Geometrical Lines, Figures, and 

Curves. By E. W. Tarn, M.A., Architect. 8vo, cloth $3.50 

"No book with the same objects in view has ever been published in which 

the clearness of the rules laid down and the illustrative diagrams have been 

so satisfactory." — Scotsman. 

THE GEOMETRY OF COMPASSES. 

Or, Problems Resolved by the mere Description of Circles and the Use 
of Coloured Diagrams and Symbols. By Oliver Byrne. Coloured 
Plates. 12mo, cloth $1.50 

MENSURATION AND MEASURING. 

With the Mensuration and Levelling of Land for the purposes of Modern 
Engineering. By T. Baker, C.E. New Edition by E. Nugent, C.E. 
12mo, cloth # (3Q 

HANDY BOOK FOR THE CALCULATION OF STRAINS 

In Girders and Similar Structures and their Strength. Consisting of 
Formulae and Corresponding Diagrams, with numerous details for Prac- 
tical Application, &c. By William Humber, A.M.Inst.C.E., &c. Sixth 
Edition. 12mo, with nearly 100 Woodcuts and 3 Plates, cloth. $2.50 

THE STRAINS ON STRUCTURES OF IRONWORK. 

With Practical Remarks on Iron Construction. By F. W. Shields, 
M.Inst.C.E. 8vo, cloth : $2.00 

CONSTRUCTIONAL IRON AND STEEL WORK, 

As applied to Public, Private, and Domestic Buildings. By Francis 
Campin, C.E. 12mo, cloth $1.40 



18 CROSBY LOCKWOOD & SON'S CATALOGUE. 
MATERIALS AND CONSTRUCTION. 

A Theoretical and Practical Treatise on the Strains, Designing, and 
Erection of Works of Construction. By Francis Campus, C.E. Third 

Edition. 12mo, cloth SI. 20 

"Xo better exposition of the practical application of the principles of 

construction has yet been published to our knowledge in such a cheap 

comprehensive form.''' — Building A ews. 

EXPERIMENTS ON THE FLEXURE OF BEAMS. 

Resulting in the Discovery of New Laws of Failure by Buckling. By 
Albert E. Gey. Medium 8vo, cloth Xet Jjjl.^o 

TRUSSES OF WOOD AND IRON. 

Practical Applications of Science in Determining the Stresses, Breaking 
Weights, Safe Loads, Scantlings, and Details of Construction. With 
Complete Working Drawings. By W. Griffiths, Surveyor. Oblong, 

8vo, cloth SI. 80 

"This handy little book enters so minutely into every detail connected 

with the construction of roof trusses that no student need be ignorant of 

these matters." — Practical Engineer. 

CONSTRUCTION OF ROOFS, OF WOOD AND IRON: 

Deduced chieflv from the Works of Robison, Tredgold, and Humber. 

By E. W. Tarn, M.A., Architect. Fourth Edition. 12mo, cloth. .60 
"Mr. Tarn i3 so thoroughly master of his subject, that although the trea- 
tise was founded on the works of others he has given it a distinct value of 
his own. It will be found valuable by all students." — Builder. 

A TREATISE ON THE STRENGTH OF MATERIALS. 

With Rules for Application in Architecture, the Construction of Sus- 
pension Bridges, Railwavs, <fcc. By Peter Barlow, F-R-S. A new 
Edition, revised bvhis Sons. P. W. Barlow, F.R.S., and "ft . H. Barlow. 
F.R.S-; to which are added. Experiments by Hodgkixsox, Fairbairx, 
and Kirkaldy; and Formulae for calculating Girders, &c. Edited by 
Wm. Hitmber, A.M. Inst. C.E. Svo, 400 pp., with 19 Plates and numer- 
ous Woodcuts, cloth S7 .00 

"Valuable alike to the student, tyro, and the experienced practitioner, it 

will always rank in future as it has hitherto done, as the standard treatise 

on that particular subject." — Engineer. 

EXPANSION OF STRUCTURES BY HEAT. 

By John" Kelly, C.E., late of the Indian Public Works Department. 
12mo, cloth SI. 50 

"The aim the author has set before him, viz., to show the effects of heat 
upon metallic and other structures, is a laudable one, for this is a branch ot 
physics upon which the engineer or architect can find but little reliable and 
comprehensive data in books." — Builder. 

CIVIL ENGINEERING. 

By Henry Law, M.Inst. C.E. Including a Treatise on Hydraulic En- 
gineering bv G. R. Burxell. M.Inst. C.E. Seventh Edition, re vised, 
with Large" Additions on Recent Practice by D. Kixnear Clark 
M.Inst. C.E. 12mo, cloth. : S3. 60 

GAS WORKS, t jTv . 

Their Construction and Arrangement, and the Manufacture and Distri- 
bution of Coal Gas. Bv S. Hughes, C.E. Ninth Edition. Revised, 
with Notices of Recent Improvements by Henry O'Connor, AAI.Lnst. 

CE. 12mo, cloth S2.40 

"Of infinite service alike to manufacturers, distributors, and consumers." 



MARINE ENGINEERING, NAVIGATION, &>c. 19 
PNEUMATICS, 

Including Acoustics and the Phenomena of Wind Currents, for the use 
of Beginners. By Charles Tomlinson, F.R.S. 12mo, cloth. t QQ 

FOUNDATIONS AND CONCRETE WORKS. 

With Practical Remarks on Footings, Planking, Sand, Concrete, B^ton, 
Pile-driving, Caissons, and Cofferdams. By E. Dobson. 12mo. # gQ 

BLASTING AND QUARRYING OF STONE, 

For Building and other Purposes. With Remarks on the Blowing up of 
Bridges. By Gen. Sir J. Burgoyne, K.C.B. 12mo, cloth. .. . t QQ 

SAFE RAILWAY WORKING. 

A Treatise on Railway Accidents, their Cause and Prevention; with a 
Description of Modern Appliances and Systems. By Clement E. 

Stretton, C.E. Third Edition, Enlarged. 12mo, cloth $1.50 

"A book for the engineer, the directors, the managers; and, in short, all 

who wish for information on railway matters will find a perfect encyclopaedia 

in 'Safe Railway Working.'" — Railway Review. 



MARINE ENGINEERING, SHIPBUILDING, 
NAVIGATION, ETC. 



MARINE ENGINES AND BOILERS. 

Their Design and Construction. A Handbook for the Use of Students, 
Engineers, and Naval Constructors. Based on the Work "Berechnung 
und Konstruktion der Schiffsmaschinen und Kessel," by Dr. G. Bauer, 
Engineer-in-Chief of the Vulcan Shipbuilding Yard, Stettin. Translated 
from the Second German Edition by E. M. Donkin, and S. Brian 
Donkin, A.M. I. C.E. Edited by Leslie S. Robertson, Secretary to 
the Engineering Standards Committee, M.I.C.E., M.I.M.E., M.I.N.A., 
&c. With numerous Illustrations and Tables. Thiek 8vo, cloth, 

[Just Published. $9,00 
Summary of Contents:— PART I.— MAIN ENGINES.— Determina- 
tion op Cylinder Dimensions. — The Utilisation of Steam in the En- 
gine. — Stroke of Piston. — Number of Revolutions. — Turning Moment. 
— Balancing of the Moving Parts. — Arrangement of Main Engines. — 
Details of Main Engines. — The Cylinder. — Valves. — Various Kinds 
of Valve Gear. — Piston Rods. — Pistons. — Connecting Rod and Cross- 
head. — Valve Gear Rods. — Bed Plates. — Engine Columns. — Revers- 
ing and Turning Gear. PART II. — PUMPS. — Air, Circulating Feed 
and Auxiliary Pumps. PART III.— SHAFTING, RESISTANCE OF 
SHIPS, PROPELLERS.— Thrust Shaft and Thrust Block.— Tunnel- 
Shafts and Plummer Blocks. — Shaft Couplings. — Stern Tube. — The 
Screw Propeller. — Construction of the Screw. PART IV. — PIPES 
AND CONNECTIONS.— General Remarks, Flanges, Valves, &c— 
Under Water Fittings. — Main Steam, Auxiliary Steam, and Exhaust 
Piping. — Feed Water, Bilge, Ballast and Circulating Pipes. PART 
V. — STEAM BOILERS. — Firing and the Generation of Steam. — 
Cylindrical Boilers. — Locomotive Boilers. — Water-tube Boilers. — 
Small Tube Water-Tube Boilers. — Smoke Box. — Funnel and Boiler 
Lagging. — Forced Draught. — Boiler Fittings and Mountings. 
PART VI.— MEASURING INSTRUMENTS. PART VII.— VARIOUS 
DETAILS. — Bolts, Nuts, Screw Threads, &c. — Platforms, Gratings, 
Ladders. — Foundations. — Seatings. — Lubrication. — Ventilation of 
Engine Rooms. — Rules for Spare Gear. PART VIII. — ADDITIONAL 
TABLES. 



20 CROSBY LOCKWOOD & SON'S CATALOGUE. 
THE NAVAL ARCHITECT'S AND SHIPBUILDER'S 

POCKET-BOOK 

Of Formulae, Rules, and Tables, and Marine Engineer's and Surveyor's 
Handy Book of Reference. By Clement Mackrow, M.I.N. A. Eighth 
Edition, carefully Revised and Enlarged. Ecap, leather. . Ael So.00 
Summary of Contents:— Signs and Symbols, Decimal Fractions. — 
Trigonometry. — Practical Geometry. — Mensuration. — Centres and 
Moments of Figures. — Moments of Inertia and Radii Gyration — Al- 
gebraical Expressions for Simpson's Rules. — Mechanical Principles. 
— Centre of Gravity. — Laws of Motion. — Displacement, Centre of 
Buoyancy. — Centre of Gravity of Ships' Hull. — Stability Curves and 
Metacentres. — Sea and Shallow- water Waves. — Rolling of Ships. — 
Propulsion and Resistance of Vessels. — Speed Trials. — Sailing, Cen- 
tre of Effort. — Distances down Rivers, Coast Lines. — Steering and 
Rudders of Vessels. — Launching Calculations and Velocities. — 
Weight of Material and Gear. — Gun Particulars and Weight. — 
Standard Gauges. — Riveted Joints and Riveting. — Strength and 
Tests of Materials. — Binding and Shearing Stresses. — Strength of 
Shafting, Pillars, Wheels, &c. — Hydraulic Data, &c. — Conic Sec- 
tions, Catenarian Curves. — Mechanical Powers, Work. — Board of 
Trade Regulations for Boilers and Engines. — Board of Trade Reg- 
ulations for Ships. — Lloyd's Rules for Boilers. — Lloyd's Weight of 
Chains. — Lloyd's Scantlings for Ships. — Data of Engines and Ves- 
sels. — Ships' Fittings and Tests. — Seasoning Preserving Timber. — ■ 
Measurement of Timber. — Alloys, Paints, Varnishes. — Data for Stow- 
age. — Admiralty Transport Regulations. — Rules for Horse-power, 
Screw Propellers, &c. — Percentages for Butt Straps. — Particulars 
of Yachts. — Masting and Rigging. — Distances of Foreign Ports. — 
Tonnage Tables. — Vocabulary of French and English Terms. — English 
Weights and Measures. — Foregn Weights and Measures. — Decimal 
Equivalents. — Useful Numbers. — Circular Measures. — Areas of and 
Circumferences of Circles. — Areas of Segments of Circles. — Tables 
of Squares and Cubes and Roots of Numbers. — Tables of Logarithms 
of Numbers. — Tables of Hyberpolic Logarithms. — Tables of Natural 
Sines, Tangents. — Tables of Logarithmic Sines, Tangents, &c. 

WANNAN'S MARINE ENGINEER'S GUIDE 

To Board of Trade Examinations for Certificates of Competency. Con- 
taining all Latest Questions to Date, with Simple, Clear, and Correct 
Solutions; 302 Elementary Questions with Illustrated Answers, and 
Verbal Questions and Answers; complete Set of Drawings with State- 
ments completed. By A. C. Wannan, C.E., Consulting Engineer, and 
E. W. I. Wannan, M.I.M.E., Certificated First Class Marine Engineer. 
With numerous Engravings. Fourth Edition, Enlarged. 500 pages. 
8vo, cloth S4.00 

WANNAN'S MARINE ENGINEER'S POCKET=BOOK. 

Containing Latest Board of Trade Rules and Data for Marine Engineers. 
By A. C. Wannan. Third Edition, Revised, Enlarged, and Brought up 
to Date. Square 18mo, with thumb Index, leather S2.00 

MARINE ENGINES AND STEAM VESSELS. 

By R. Murray, C.E. Eighth Edition, thoroughly Revised, with Addi- 
tions by the Author and by George Carlisle, C.E. 12mo, cloth . SI ,80 

ELEMENTARY MARINE ENGINEERING. 

A Manual for Young Marine Engineers and Apprentices. By J. S. 
Brewer. 12mo, cloth .60 

CHAIN CABLES AND CHAINS. 

Comprising Sizes and Curves of Links, Studs, &c, Iron for Cables and 
Chains, Chain Cable and Chain Making, Forming and Welding Links, 
Strength of Cables and Chains, Certificates for Cables, Marking Cables, 
Prices of Chain Cables and Chains, Historical Notes, Acts of Parlia- 
ment, Statutory Tests, Charges for Testing, List of Manufacturers of 



MARINE ENGINEERING, NAVIGATION, &c 21 

Cables, &c, &c. By Thomas W, Traill, F.E.R.N., M.Inst.C.E., En- 
gineer-Surveyor-in-Chief , Board of Trade, Inspector of Chain Cable and 
Anchor Proving Establishments, and General Superintendent, Lloyd's 
Committee on Proving Establishments. With numerous Tables, Illus- 
trations, and Lithographic Drawings. Folio, cloth $15.00 

THE SHIPBUILDING INDUSTRY OF GERMANY. 

Compiled and Edited by G. Lehmann-Felskowski. With Coloured 
Prints, Art Supplements, and numerous Illustrations throughout the 
text. Super-royal 4to, cloth $4.20 

SHIPS AND BOATS. 

By W. Bland. With numerous Illustrations and Models. Tenth Edi- 
tion. 12mo, cloth .60 

SHIPS FOR OCEAN AND RIVER SERVICE, 

Principles of the Construction of. By H. A. Sommerfeldt. 12mo. 

.60 
AN ATLAS OF ENGRAVINGS 

To illustrate the above. Twelve large folding Plates. Royal 4to, 

cloth $3.00 

NAVAL ARCHITECTURE. 

An Exposition of the Elementary Principles. By J. Peake. 12mo. 
doth $1.40 

THE ART AND SCIENCE OF SAILMAKING. 

By Samuel B. Sadler, Practical Sailmaker, late in the employment of 
Messrs. Ratsey and Lapthorne, of Cowes and Gosport. Plates. 4to, 

doth $5.00 

"This extremely practical work gives a complete education in all the 
branches of the manufacture, cutting out, roping, seaming, and goring. It 
s copiously illustrated, and forms a first-rate text-book and guide." 

SAILS AND SAIL=MAKING. 

With Draughting, and the Centre of Effort of the Sails. Weights and 
Sizes of Ropes; Masting, Rigging, and Sails of Steam Vessels, &c. "By 
R. Kipping, N.A. 12mo, cloth $1.00 

MASTING, MAST=MAKING, AND RIGGING OF SHIPS. 

Also Tables of Spars, Rigging, Blocks; Chain, Wire, and Hemp Ropes, 
&c, relative to every class of vessels. By R. Kipping. 12mo, cloth, 

.80 
SEA TERMS, PHRASES, AND WORDS 

(Technical Dictionary of) used in the English and French Languages 
(English- French, French-English). For the Use of Seamen, Engineers, 
Pilots, Shipbuilders, Shipowners, and Ship-brokers. Compiled by W. 
Pirrie, late of the African Steamship Company. Fcap, 8vo, cloth 

li m P $2.00 

This volume will be highly appreciated by seamen, engineers, pilots, ship- 
builders and shipowners. It will be found wonderfully accurate and com- 
plete. 

SAILOR'S SEA BOOK: 

A Rudimentary Treatise on Navigation. By James Greenwood, B.A. 
With numerous Woodcuts and Coloured Plates. New and Enlarged 

Edition. By W. H. Rosser. 12mo, cloth $1.00 

Is perhaps the best and simplest epitome of navigation ever compiled. 

PRACTICAL NAVIGATION. 

Consisting of the Sailor's Sea Book, by J. Greenwood and W. H. Rosser; 
together with Mathematical and Nautical Tables for the Working of the 
Problems, by H. Law, C.E., and Prof. J. R. Young $2 8Q 



22 CROSBY LOCKWOOD & SOX'S CATALOGUE. 
NAVIGATION AND NAUTICAL ASTRONOMY, 

In Theory and Practice. By Prof. J. R. Young. 12mo, cloth.Sl.OO 
"A very complete, thorough, and useful manual for the young navigator." 

MATHEMATICAL TABLES, 

For Trigonometrical, Astronomical, and Nautical Calculations; to 
which is prefixed a Treatise on Logarithms, by H. Law, C.E. With 
Tables for Navigation and Nautical Astronomy. By Prof. J. R. Yocng. 
12mo, cloth SI. 60 



MINING, METALLURGY, AND 
COLLIERY WORKING. 



THE OIL FIELDS OF RUSSIA AND THE RUSSIAN 

PETROLEUM INDUSTRY. . . 

A Practical Handbook on the Exploration, Exploitation, and Manage- 
ment of Russian Oil Properties, including Notes on the Origin of Petro- 
leum in Russia, a Description of the Tneory and Practice of Liquid 
Fuel, and a Translation of the Rules and Regulations concerning Rus- 
sian Oil Properties. By A. Beebt Thompson. A.M.I.M.E., late Chief 
Engineer and Manager of the European Petroleum Company's Russian 
Oil Properties. About 500 pp. With numerous Illustrations and 
Photographic Plates, and a Map of the Balakhany-Saboontchy-Romany 
Oil Field. Royal 8vo, cloth Net § 7 ,50 

MECHANICS OF AIR MACHINERY. 

By Dr.' J. Weisbach and Prof. G. Heebmaxx. Authorized Translation 
with an Appendix on American Practice by A. Trowbridge, Ph.B., 
Adjunct Professor of Mechanical Engineering, Columbia University. 

Royal 8vo, cloth [Just Published. Xet §3.75 

Summary of Contents: — The Movement of Air. — Natcral and Arti- 
ficial Ventilation. — Blowing-Engines; — Vacuum Pumps; Tcteres; 
Hot-air Blast; Work Performed bt Blowers; Blast-reservoirs; 
Piston-blowers. — Compressors. — Rotary Blowers. — Fans. — Recent 
American Practice, d:c. 

MACHINERY FOR METALLIFEROUS MINES. 

\ Practical Treatise for Mining Engineers. Metallurgists, and Managers 
of Mines. Bv E. Henry Davies, M.E.. F.G.S. 600 pp. With Fold- 
ing Plates and other Illustrations. Medium 8vo, cloth 88.00 

'Deals exhaustively with the many and complex details which go to 
make up the sum total of machinery and other requirements for the success- 
ful working of metalliferous mines, and as a book of ready reference is of 
the highest value to mine managers and directors." — Mining Journal. 

THE DEEP LEVEL MINES OF THE RAND, 

And their Future Development, considered from the Commercial Point 
of View. By G. A. Denny (of Johannesburg), M.N.E.I.M.E.. Con- 
sulting Engineer to the General Mining and Finance Corporation, Ltd., 
of London. Berlin. Paris, and Johannesburg. Fully Illustrated with 

Diagrams and Folding Plates. Royal 8vo, buckram §10.00 

''Mr. Denny by confining himself to the consideration of the future of the 
deep-level mines" of the Rand breaks new ground, and by dealing with the 
subject rather from a commercial standpoint than from a scientific one, 
appeals to a wide circle of readers. The book cannot fail to prove of very 
great value to investors in South African mines." — Mining Journal. 



MINING, METALLURGY, & COLLIERY WORKING. 23 
PROSPECTING FOR GOLD. 

A Handbook of Practical Information and Hints for Prospectors based 
on Personal Experience. By Daniel J. Rankin, F.R.S.G.S., M.R.A.S., 
formerly Manager of the Central African Company, and Leader of 
African Gold Prospecting Expeditions. With Illustrations specially 

Drawn and Engraved for the Work. Fcap. 8vo, leather $3.00 

" This well-compiled book contains a collection of the richest gems of use- 
ful knowledge for the prospector's benefit. A special table is given to 
accelerate the spotting at a glance of minerals associated with gold." — Min- 
ing Journal. 

THE METALLURGY OF GOLD. 

A Practical Treatise on the Metallurgical Treatment of Gold-bearing 
Ores. Including the Assaying, Melting, and Refining of Gold. By M. 
Eissler, M.Inst.M.M. Fifth Edition, Enlarged. With over 300 Illus- 
trations and numerous Folding Plates. Medium 8vo, cloth. . .$7.50 
" This book thoroughly deserves its title of a 'Practical Treatise.' The 
whole process of gold mining, from the breaking of the quartz to the assay 
of the bullion, is described in clear and orderly narrative and with much 
fulness of detail." 

THE CYANIDE PROCESS OF GOLD EXTRACTION, 

And its Practical Application on the Witwatersrand Gold Fields and 
elsewhere. By M. Eissler, M.Inst.M.M. With Diagrams and Work- 
ing Drawings, Third Edition, Revised and Enlarged. 8vo, cloth, 

$3.00 

"This book is just what was needed to acquaint mining men with the 
actual working of a process which is not only the most popular, but is, as 
a general rule, the most successful for the extraction of gold from tailings." 
— Mining Journal. 

DIAMOND DRILLING FOR GOLD & OTHER MINERALS. 

A Practical Handbook on the Use of Modern Diamond Core Drills in 
Prospecting and Exploiting Mineral-Bearing Properties, including Par- 
ticulars of the Costs of Apparatus and Working. By G. A. Denny, 
M.N.E.Inst.M.E., M.Inst.M.M. Medium 8vo, 168 pp., with Illustra- 
tive Diagrams $5.00 

"There is certainly scope for a work on diamond drilling, and Mr. Denny 

deserves grateful recognition for supplying a decided want." — Mining 

Journal. 



GOLD ASSAYING. 

A Practical Handbook, giving the Modus Operandi for the Accurate 
Assay of Auriferous Ores and Bullion, and the Chemical Tests required 
in the Processes of Extraction by Amalgamation, Cyanidation, and 
Chlorination. With an Appendix of Tables and Statistics. By H. 
Joshua Phillips, F.I.C., F.C.S., Assoc Inst.C.E., Author of "Engineer- 
ing Chemistry," <fec. With numerous Illustrations. 12mo, cloth. $3,00 

FIELD TESTING FOR GOLD AND SILVER. 

A Practical Manual for Prospectors and Miners. By W. H. Merritt 
M.N.E.Inst.M.E., A.R.S.M., &c. With Photographic Plates and other 

Illustrations. Fcap. 8vo, leather SI. 50 

'As an instructor of prospectors' classes Mr. Merritt has the advantage of 
knowing exactlv the information likely to be most valuable to the miner 
in the field- The contents cover all the details for sampling and testing 
gold and silver ores. A useful addition to a prospector's kit." — Mining 
Journal. 



24 CROSBY LOCKWOOD &» SON'S CATALOGUE. 
THE PROSPECTOR'S HANDBOOK. 

A Guide for the Prospector and Traveller in search of Metal-Bearing or 
other Valuable Minerals. By J. W. Anderson, M.A.(Camb.). F.R.G.S. 

Tenth Edition. 12mo, cloth $1.50 

"How to find commercial minerals, and how to identify them when they 

are found, are the leading points to which attention is directed." — Mining 

Journal. 

THE METALLURGY OF SILVER. 

A Practical Treatise on the Amalgamation, Roasting, and Lixiviation 
of Silver Ores. Including the Assaying, Melting, and Refining of Silver 
Bullion. By M. Eissler, M.Inst.M.M. Fifth Edition. 12mo, cloth. 

$4.00 

"A practical treatise and a technical work which we are convinced will 
supply a long-felt want amongst practical men, and at the same time be of 
value to students and others indirectly connected with the industries." — 
Mining Journal. 

THE HYDRO=METALLURGY OF COPPER. 

Being an Account of Processes Adopted in the Hydro-Metallurgical 
Treatment of Cupriferous Ores, Including the Manufacture of Copper 
Vitriol, with Chapters on the Sources of Supply of Copper and the 
Roasting of Copper Ores. By M. Eissler, M.Inst.M.M. 8vo, cloth, 

$4.50 

"In this volume the various processes for the extraction of copper by wet 
methods are fully detailed. Costs are given when available, and a great 
deal of useful information about the copper industry of the world is pre- 
sented in an interesting and attractive manner." — Mining Journal. 

THE METALLURGY OF ARGENTIFEROUS LEAD. 

A Practical Treatise on the Smelting of Silver-Lead Ores and the Refin- 
ing of Lead Bullion. Including Reports on various Smelting Estab- 
lishments and Descriptions of Modern Smelting Furnaces and Plants 
in Europe and America. By M. Eissler, M.Inst.M.M. 12mo, cloth 

$5.00 

"The numerous metallurgical processes, which are fully and extensively 

treated of, embrace all the stages experienced in the passage of the lead 

from the various natural states to its issue from the refinery as an article 

of commerce." — Practical Engineer. 

METALLIFEROUS MINERALS AND MINING. 

By D. C. Da vies, F.G.S. Sixth Edition, thoroughly Revised and much 
Enlarged by his Son, E. Henry Davies, M.E., F.G.S. 600 pp., with 

173 Illustrations. 8vo, cloth Net $5.00 

"Neither the practical miner nor the general_ reader, interested in mines, 
can have a better book for his companion and his guide." — Mining Journal. 

EARTHY AND OTHER MINERALS AND MINING. 

By D. C. Davies, F.G.S., Author of "Metalliferous Minerals," &c. 
Third Edition, Revised and Enlarged by his Son, E. Henry Davies, 
M.E., F.G.S. With about 100 Illustrations. 12mo, cloth $5.00 

BRITISH MINING. 

A Treatise on the Historv, Discovery, Practical Development, and 
Future Prospects of Metalliferous Mines in the United Kingdom. By 
Robert Hunt, F.R.S., late Keeper of Mining Records. Upwards of 
950 pp., with 230 Illustrations. Second Edition, Revised. Super-royal 
8vo, cloth $15.00 

POCKET=BOOK FOR MINERS AND METALLURGISTS. 

Comprising Rules, Formulae, Tables, and Notes for Use in Field and 
Office Work. By F. Danvers Power, F.G.S., M.E. Second Edition, 
Corrected. Fcap. 8vo, leather $3.50 



MINING, METALLURGY, & COLLIERY WORKING. 25 
THE MINER'S HANDBOOK. 

A Handy Book of Reference on the subjects of Mineral Deposits, Mining 
Operations, Ore Dressing, &c. For the Use of Students and others in- 
terested in Mining Matters. Compiled by John Milne, F.R.S., Pro- 
fessor of Mining in the Imperial University of Japan. Third Edition. 
Fcap. 8vo, leather $3.00 

IRON ORES of GREAT BRITAIN and IRELAND. 

Their Mode of Occurrence, Age and Origin, and the Methods of Searching 
for and Working Them. With a Notice of some of the Iron Ores of 
Spain. By J. D. Kendall, F.G.S., Mining Engineer. 12mo, cloth 

$6.00 
METALLURGY OF IRON. 

Containing History of Iron Manufacture, Methods of Assay, and Analy- 
ses of Iron Ores, Processes of Manufacture of Iron and Steel, &c. By 
H. Bauerman, F.G.S., A.R.S.M. With numerous Illustrations. Sixth 

Edition, revised and enlarged. 12mo, cloth $2.00 

"Carefully written, it has the merit of brevity and conciseness, as to less 

important points ; while all material matters are very fully and thoroughly 

entered into." — Standard. 

MINE DRAINAGE. 

A Complete Practical Treatise on Direct-Acting Underground Steam 
Pumping Machinery. By Stephen Michell. Second Edition, Re- 
written and Enlarged. With 250 Illustrations. Royal 8vo, cloth. 

$10 00 

HORIZONTAL PUMPING ENGINES.— Rotary and Non-Rot'art 
Horizontal Engines. — Simple and Compound Steam Pumps. — VERTI- 
CAL PUMPING ENGINES.— Rotary and Non-Rotary Vertical 
Engines. — Simple and Compound Steam Pumps. — Triple-Expansion 
Steam Pumps. — Pulsating Steam Pumps. — Pump Valves. — Sinking 
Pumps, &c., &c. 

ELECTRICITY AS APPLIED TO MINING. 

By Arnold Lupton, M.Inst.C.E., M.I.M.E., M.I.E.E., late Professor of 
Coal Mining at the Yorkshire College, Victoria University, Mining En- 
gineer and Colliery Manager; G. D. Aspinall Parr, M.I.E.E., A.M.I. 
M.E., Associate of the Central Technical College, City and Guilds of 
London, Head of the Electrical Engineering Department, Yorkshire 
College, Victoria University; and Herbert Perkin, M.I.M.E., Certifi- 
cated Colliery Manager, Assistant Lecturer in the Mining Department of 
the Yorkshire College, Victoria University. With about 170 Illustra- 
tions. Second Edition, Revised and Enlarged. Medium 8vo, cloth. 

[Just Published.] $4,50 
(For Summary of Contents, see page 28.) 

THE COLLIERY MANAGER'S HANDBOOK. 

A Comprehensive Treatise on the Laying-out and Working of Collieries, 
Designed as a Book of Reference for Colliery Managers, and for the 
Use of Coal-Mining Students preparing for First-class Certificates. By 
Caleb Pamely, Mining Engineer and Surveyor; Member of the North, 
of England Institute of Mining and Mechanical Engineers; and Member 
of the South Wales Institute of Mining Engineers. With over 1,000 
Diagrams, Plans, and other Illustrations. Fifth Edition, Carefully 
Revised and Greatly Enlarged. 1,200 pp. Medium 8vo, cloth. $10.00 
Geology. — Search for Coal. — Mineral Leases and other Holdings. — 
Shaft Sinking. — Fitting Up the Shaft and Surface Arrangements. — 
Steam Boilers and their Fittings. — Timbering and Walling. — Narrow 
Work and Methods of Working. — Underground Conveyance. — Drain- 
age. — The Gases met with in Mines; Ventilation. — On the Friction of 
Air in Mines. — The Priestman Oil Engine; Petroleum and Natural 
Gas. — Surveying and Planning. — Safety Lamps and Firedamp Detect- 
ors. — Sundry and Incidental Operations and Appliances. — Colliery 
Explosions. — Miscellaneous Questions and Answers. — Appendix: 
Summary of Report of H.M. Commissioners on Accidents in Mines. 



26 CROSBY LOCKWOOD &> SON'S CATALOGUE. 
PRACTICAL COAL=MINING. 

An Elementary Class-Book for the Use of Students attending Classes in 
Preparation for the Board of Education and County Council Examina- 
tions, or Qualifying for First or Second Class Colliery Managers' Cer- 
tificates. By. T. H. Cockin, Member of the Institution of Mining 
Engineers, Certificated Colliery Manager, Lecturer on Coal-Mining at 
Sheffield University College. With Map of the British Coal-fields and 
over 200 Illustrations specially Drawn and Engraved for the Work. 
440 pp., 12mo, cloth $2.50 

COLLIERY WORKING AND MANAGEMENT. 

Comprising the Duties of a Colliery Manager, the Oversight and Arrange- 
ment of Labour and Wages, and the different Systems of Working Coal 
Seams. By H. F. Bulman and R. A. S. Redmayne. 350 pp., with 
28 Plates and other Illustrations, including Underground Photographs. 
Medium 8vo, cloth Net $6.00 

NOTES AND FORMULAE FOR MINING STUDENTS. 

By John Herman Merivale, M.A., Late Professor of Mining in the 
Durham College of Science, Newcastle-upon-Tyne. Fourth Edition, 
Revised and Enlarged. By H. F. Bulman, A.M.Inst. C.E. 12mo, 

cloth $1.00 

"The author has done his work in a creditable manner, and has produced 

a book that will be of service to students and those who are practically 

engaged in mining operations." — Engineer. 

PHYSICS AND CHEMISTRY OF MINING. 

An Elementary Class-Book for the use of Students preparing for the 
Board of Education and County Council Examinations in Mining, or 
Qualifying for Colliery Managers' Certificates. By T. H. Byrom, 
Chemist to the Wigan Coal and Iron Co-, Ltd., &c. With Illustrations. 
12mo, cloth [Just Published.] $1.50 

MINING CALCULATIONS. 

For the use of Students Preparing for the Examinations for Colliery 
Managers' Certificates, comprising Numerous Rules and Examples in 
Arithmetic, Algebra, and Mensuration. By T. A. O'Donahue, M.E., 
First-class Certificated Colliery Manager. 12mo, cloth $1.50 

COAL AND COAL MINING. 

By the late Sir Warington W. Smyth, M.A., F.R.S. Eighth Edition, 
Revised and Extended by T. Forster Brown, Chief Inspector of the 
Mines of the Crown and of the Duchy of Cornwall. 12mo, cloth. $1.40 

INFLAMMABLE GAS AND VAPOUR IN THE AIR 

(The Detection and Measurement of). By Frank Clowes, D.Sc, 
Lond., F.I.C. With a Chapter on The Detection and Measurement 
of Petroleum Vapour, by Boverton Redwood, F.R.S.E. 12mo, 

cloth $2.50 

"Professor Clowes has given us a volume on a subject of much industrial 
importance. . . Those interested in these matters may be recommended 
to study this book, which is easy of comprehension and contains many good 
things." — The Engineer. 

COAL & IRON INDUSTRIES of the UNITED KINGDOM. 

Comprising a Description of the Coal Fields and of the Principal Seams 
of Coal, with Returns of their Produce and its Distribution, and Analyses 
of Special Varieties. Also, an Account of the Occurrence of Iron Ore 
in Veins or Seams; Analyses of each Variety; and a History of the 
Rise and Progress of Pig Iron Manufacture. By Richard Meade. 8vo, 
Cloth $10.00 



MINING, METALLURGY, & COLLIERY WORKING. 27 
MINING TOOLS, 

Manual of. By W. Morgans, Lecturer on Mining at the Bristol School 

of Mines. 12mo, cloth $1.00 

Atlas of Engravings to the above, containing 235 Illustrations drawn 
to Scale. 4to $1.80 

SLATE AND SLATE QUARRYING. 

Scientific, Practical, and Commercial. By D. C. Davies, F.G.S., Min- 
ing Engineer, &c. With numerous Illustrations and Folding Plates 
Fourth Edition. 12mo, cloth : $1.20 

A FIRST BOOK OF MINING AND QUARRYING. 

By J. H. Collins, F.G.S. Crown 8vo, cloth .60 

ASBESTOS AND ASBESTIC. 

Their Properties, Occurrence, and Use. By Robert H. Jones, F.S.A. 
Mineralogist, Hon. Mem. Asbestos Club, Black Lake, Canada. With 
Ten Collotype Plates and other Illustrations. Demy 8vo, cloth. $6.40 

GRANITES AND OUR GRANITE INDUSTRIES. 

By George F. Harris, F.G.S. With Illustrations. 12mo, cloth. $1,00 

MINERAL SURVEYOR AND VALUER'S GUIDE. 

Comprising a Treatise on Improved Mining Surveying and the Valuation 
of Mining Properties, with New Traverse Tables. By W. Lintern, C.E. 

Fourth Edition, enlarged. 12mo, cloth $1.40 

"Contains much valuable information, and is thoroughly trustworthy."— 
Iron and Coal Trades Review. 

TRAVERSE TABLES. 

For use in Mine Surveying. By William Lintern, C.E. With two 
plates. Small crown 8vo, cloth Net $1.50 

SUBTERRANEOUS SURVEYING. 

By T. Fenwick. Also the Method of Conducting Subterraneous Sur- 
veys without the use of the Magnetic Needle, &c. By T.Baker. 12mo, 

$1.00 
MINERALOGY, 

Rudiments of. By A. Ramsay, F.G.S. Fourth Edition. Woodcuts 
and Plates. 12mo, cloth $1.40 

PHYSICAL GEOLOGY, 

Partly based on Major-General Portlock's "Rudiments of Geology." 
By Ralph Tate, A.L.S., &c. Woodcuts. 12mo, cloth .80 

HISTORICAL GEOLOGY, 

Partly based on Major-General Portlock's "Rudiments." By Ralph 
Tate. 12mo, cloth $1.00 

GEOLOGY, 

Physical and Historical. Consisting of "Physical Geology," which 
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ogy," which treats of the Mineral and Organic Conditions of the Earth 
at each successive epoch. By R. Tate. 12mo, cloth $1.80 



2% CROSBY LOCKWOOD & SON'S CATALOGUE. 



ELECTRICITY, ELECTRICAL 
ENGINEERING, ETC. 



THE ELEMENTS OF ELECTRICAL ENGINEERING. 

A First Year's Course for Students. _ By Tyson Sewell, A.I.E.E., 
Assistant Lecturer and Demonstrator in Electrical Engineering at the 
Polytechnic, Regent Street, London. Third Edition, Revised and En- 
larged, including an Appendix of Questions and Answers. 460 pages, 
with 274 Illustrations. Demy 8vo, cloth. . . . [Just Published.] §3.00 
Ohm's Law. — Units Employed in Electrical, Engineering. — Series, 
and Parallel Circuits; Current Density and Potential Drop in the. 
Circuit. — The Heating Effect of the Electric Current. — The Mag- 
netic Effect of an Electric Current. — The Magnetisation of Iron. — 
Electro-Chemistry ; Primary Batteries. — Accumulators. — Indicat- 
ing Instruments; Ammeters, Voltmeters, Ohmmeters. — Electricity 
Supply Meters. — Measuring Instruments, and the Measurement of 
Electrical Resistance. — Measurement of Potential Difference 
Capacity Current Strength, and Permeability. — Arc Lamps. — Incan- 
descent Lamps, Manufacture and Installation; Photometry. — The 
Continuous Current Dynamo. — Direct Current Motors. — Alternating 
Currents. — Transformers, Alternators, Synchronous Motors. — Poly- 
phase Working. — Appendix of Questions and Answers. 

ELEMENTARY ELECTRICAL ENGINEERING 

In Theory and Practice. A Class-book for Junior and Senior Students 
and Working Electricians. By J. H. Alexander, M.B., A.I.E.E 
With 181 Illustrations. 12mo, cloth [Just Published. §1.50 

THE ELECTRICAL TRANSMISSION OF ENERGY. 

A Manual for the Design of Electrical Circuits. By Arthur Vaughan 
Abbott, C.E., Member American Institute of Electrical Engineers, 
Member American Institute of Mining Engineers, Member American 
Society of Civil Engineers, Member American Society of Mechanical 
Engineers, &c. With Ten Folding Diagrams and Sixteen Full-page 
Engravings. Fourth Edition, entirely Re-Written and Enlarged. 
Royal 8vo, cloth Net §5.00 

ELECTRICITY AS APPLIED TO MINING. 

By Arnold Lupton, M.Inst.C.E., M.I.M.E., M.I.E.E., late Professor of 
Coal Mining at the Yorkshire College, Victoria University, Mining En- 
gineer and Colliery Manager; G. D. Aspinall Parr, M.I.E.E., A.M.I. 
M.E., Associate of the Central Technical College, City and Guilds of 
London, Head of the Electrical Engineering Department, Yorkshire 
College, Victoria University; and Herbert Perkin, M.I.M.E., Cer- 
tificated Colliery Manager, Assistant Lecturer in the Mining Depart- 
ment of the Yorkshire College, Victoria University. With about 170 
Illustrations. Second Edition, Revised and Enlarged. Medium 8vo, 

cloth [Just Published. §4.50 

Introductory. — Dynamic Electricity. — Driving of the Dynamo. — 
The Steam Turbine. — Distribution of Electrical Energy. — Starting 
and Stopping Electrical Generators and Motors. — Electric Cables. — 
Central Electrical Plants. — Electricity applied to Pumping and 
Hauling. — Electricity applied to Coal-cutting. — Typical Electric 
Plants Recently Erected. — Electric Lighting by Arc and Glow 
Lamps. — Miscellaneous Applications of Electricity. — Electricity as 
Compared with other modes of Transmitting Power. — Dangers of 
Electricity. 



ELECTRICITY, ELECTRICAL ENGINEERING, &c. 29 
CONDUCTORS FOR ELECTRICAL DISTRIBUTION. 

Their Materials and Manufacture, The Calculation of Circuits, Pole-Line 
Construction, Underground Working, and other Uses. By F. A. C. 
Perrine, A.M., D.Sc. ; formerly Professor of Electrical Engineering, 
Leland Stanford, Jr., University; M.Amer.I.E.E. 8vo, cloth. 

Ne * $3.50 

Conductor Materials. — Alloyed Conductors. — Manufacture of 
Wire. — Wire-Finishing. — Wire Insulation. — Cables. — Calculation of 
Circuits. — Kelvin's Law of Economy in Conductors. — Multiple Arc 
Distribution. — Alternating Current Calculation. — Overhead Lines. 
— Pole Line. — Line Insulators. — Underground Conductors. 

DYNAMO ELECTRIC MACHINERY: its CONSTRUCT 

TION, DESIGN, and OPERATION. 

By Samuel Sheldon, A.M., Ph.D., Professor of Physics and Electrical 
Engineering at the Polytechnic Institute of Brooklyn, assisted by H. 
Mason. B.S. 

In two volumes, sold separately, as follows: — 

Vol. I.— DIRECT CURRENT MACHINES. Fifth Edition, Revised. 

8vo. 280 pages, with 200 Illustrations N et f$2,50 

Vol. II— ALTERNATING CURRENT MACHINES. 8vo. 260 pages, 

with 184 Illustrations Net $£.50 

Designed as Text-books for use in Technical Educational Institutions, and 
by Engineers whose work includes the handling of Direct and Alternating 
Current Machines respectively, and for Students proficient in mathematics. 

DYNAMO, MOTOR AND SWITCHBOARD CIRCUITS 

FOR ELECTRICAL ENGINEERS. 

A Practical Book dealing with the subject of Direct, Alternating and 
Polyphase Currents. By William R. Bowker, C.E., M.E., E.E., Con- 
sulting Tramway Engineer. 8vo, cloth. . .- $2.25 

"Mr. Bowker's book consists chiefly of diagrams of connections, with short 
explanatory notes, there are over 100 diagrams, and the cases considered 
cover all the more important circuits, whether in direct current, single- 
phase, or polyphase work." — Nature. 

ARMATURE WINDINGS OF DIRECT CURRENT 

DYNAMOS. 

Extension and Application of a General Winding Rule. By E. Arnold, 
Translated from the German by F. B. De Grees. 8vo, cloth. $2,00 

POWER TRANSMITTED BY ELECTRICITY, 

And applied by the Electric Motor, including Electric Railway Con- 
struction. By P. Atkinson, A.M., Ph.D. Third Edition, Fully Re- 
vised, and New Matter added. With 94 Illustrations. 12mo, cloth. 

$2.00 
DYNAMO CONSTRUCTION. 

A Practical Handbook for the Use of Engineer-Constructors and Elec- 
trieians-in-Charge. Embracing Framework Building, Field Magnet and 
Armature Winding and Grouping, Compounding, &c. By J. W. Urqu- 
hart. Second Edition, Enlarged, with 114 Illustrations. 12mo, cloth. 

$3.00 
HOW TO MAKE A DYNAMO. 

A Practical Treatise for Amateurs. Containing Illustrations and De- 
tailed Instructions for Constructing a Small Dynamo to Produce the 
Electric Light. By Alfred Crofts. Seventh Edition. 12mo, cloth. 

.80 
WIRELESS TELEGRAPHY; 

Its Origins, Development, Inventions, and Apparatus. By Charles 
Henby Sewall. With 85 Diagrams and Illustrations. 8vo, cloth. 

Met $2.00 



30 CROSBY LOCKWOOD &> SON'S CATALOGUE. 
SUBMARINE TELEGRAPHS; 

Their History, Construction, and Working. Founded in part on Wun- 
schendorff's "Traits de Telegraphie Sous-Marine," and Compiled from 
Authoritative and Exclusive Sources. By Charles Bright, F.R.S.E. 
A.M.Inst.C.E., M.I.E.E. 780 pp., fully Illustrated, including Maps and 
Folding Plates. Royal 8vo, cloth $25.00 

ELECTRICAL AND MAGNETIC CALCULATIONS. 

For the Use of Electrical Engineers and Artisans, Teachers, Students, 
and all others interested in the Theory and Application of Electricity 
and Magnetism. By Prof. A. A. Atkinson, Ohio University. 12mo, 
cloth. . . : $1.50 

"To teachers and those who already possess a fair knowledge of their sub- 
ject we can recommend this book as being useful to consult when requiring 
data or formulae which it is neither convenient nor necessary to retain by 
memory." — The Electrician. 

THE ELECTRICAL ENGINEER'S POCKET=BOOK. 

Consisting of Rules, Formulae, Tables, and Data. By H. R. Kempe, 
M.I.E.E., A.M.Inst.C.E., Technical Officer Postal Telegraphs, Author 
of "A Handbook of Electrical Testing." Second Edition. 32mo, 
leather $1.75 

ELECTRIC LIGHTING (ELEMENTARY PRINCIPLES OF). 

By Alan A. Campbell Swinton, M.Inst.C.E., M.I.E.E. Sixth Edition. 
With 16 Illustrations. 12mo, cloth .60 

ELECTRIC LIGHT. 

Its Production and Use, Embodying Plain Directions for the Treatment 
of Dynamo-Electric Machines, Batteries, Accumulators, and Electric 
Lamps. By J. W. Urquhart, C.E. Seventh Edition. 12mo, cloth. 

$3.00 

ELECTRIC LIGHT FOR COUNTRY HOUSES. 

A Practical Handbook on the Erection and Running of Small Installa- 
tions, with Particulars of the Cost of Plant and Working. By J. H. 
Knight. Fourth Edition, Revised. 12mo, boards .50 

ELECTRIC LIGHT FITTING. 

A Handbook for Working Electrical Engineers, embodying Practical 
Notes on Installation Management. By J. W. Urqukart. With 
numerous Illustrations. Fourth Edition, Revised. 12mo, cloth. $2.00 

ELECTRIC SHIP=LIGHTING. 

A Handbook on the Practical Fitting and Running of Ships' Electrical 
Plant. For the Use of Shipowners and Builders, Marine Electricians, 
and Seagoing Engineers-in-Charge. By J. W. Urquhart, C.E. Third 
Edition, Revised and Extended. With 88 Illustrations, 12mo, 
cloth $3.00 

DYNAMIC ELECTRICITY AND MAGNETISM. 

By Philip Atkinson, A.M., Ph.D., Author of "Elements of Static 
Electricity," &c. Crown, 8vo, 417 pp., with 120 Illustrations, cloth 

$2.00 

THE STUDENT'S TEXT=BOOK OF ELECTRICITY. 

By H. M. Noad, F.R.S. 650 pp., with 470 Illustrations. 12mo, 
eloth $4.00 



ARCHITECTURE, BUILDING, &c. 31 

ARCHITECTURE, BUILDING, ETC. 



SPECIFICATIONS IN DETAIL. 

By Frank W. Macey, Architect, Author of "Conditions of Contract." 
Second Edition, Revised and Enlarged, containing 644 pp., and 2,000 

Illustrations. Royal 8vo, cloth $8.00 

Summary op Contents: — General, Notes (including Points in Speci- 
fication Writing, The Order of a Specification, and Notes on Items 
often Omitted from a Specification). — Form of Outside Cover to a 
Specification. — Specification of Works and List of General, Condi- 
tions. — Preliminary Items (including Shoring and House Breaker). 
— Drainage (including Rain-water Wells and Reports). — Excavator 
(including Concrete Floors, Roofs, Stairs, and Walls). — Pavior. — 
Bricklayer (including Flintwork, River, and other Walling, Spring- 
water Wells, Storage Tanks, Fountains, Filters, Terra Cotta and 
Faience). — Mason. — Carpenter, Joiner, and Ironmonger (including 
Fencing and Piling). — Smith and Founder (including Heating, Fire 
Hydrants, Stable and Cow-house Fittings). — Slater (including Slatb 
Mason). — Tiler. — Stone Tiler. — Shingler. — Thatcher. — Plumber (in- 
cluding Hot-water Work). — Zincworker. — Coppersmith. — Plasterer. 
— Gasfitter. — Bellhanger. — Glazier. — Painter. — Paperhanger. — 
General Repairs and Alterations. — Ventilation. — Road-making. — 
Electric Light. — Index. 

PRACTICAL BUILDING CONSTRUCTION. 

A Handbook for Students Preparing for Examinations, and a Book 
of Reference for Persons Engaged in Building. By John Parnell 
Allen, Surveyor, Lecturer on Building Construction at the Durham 
College of Science, Newcastle-on-Tyne. Fourth Edition, Revised and 
Enlarged. Medium 8vo, 570 pp., with over 1,000 Illustrations, cloth, 

$3.00 
SPECIFICATIONS FOR PRACTICAL ARCHITECTURE. 

A Guide to the Architect, Engineer, Surveyor, and Builder. Upon 
the Basis of the Work by A. Bartholomew, Revised, by F. Rogers. 
8vo, cloth $6.00 

SCIENCE OF BUILDING: 

An Elementary Treatise on the Principles of Construction. By E. 
Wyndham Tarn, M.A.Lond. Fourth Edition. 12mo, cloth. $1.40 

ART OF BUILDING, 

Rudiments of. General Principles of Construction, Character, Strength, 
and Use of Materials, Preparation of Specifications and Estimates, &c. 
By Edward Dobson, M.Inst.C.E. Fifteenth Edition, revised by J. P. 
Allen, Lecturer on Building Construction at the Durham College of 
Science. 12mo, cloth .80 

BOOK ON BUILDING, 

Civil and Ecclesiastical. By Sir Edmund Beckett, Bart., LL.D. 
Second Edition. 12mo, cloth $1.80 

BUILDING ESTATES: 

A Treatise on the Development, Sale, Purchase, and Management of 
Building Land. By F. Maitland. Fourth Edition. 12mo, cloth, 

.80 
COTTAGE BUILDING) 

By C Bruce Allen. Twelfth Edition, with Chapter on Economic 
Cottages for Allotments by E. E. Allen, C.E. 12mo, cloth. . .80 



32 CROSBY LOCKWOOD &= SOX'S CATALOGUE. 
DWELLING=HOUSES, 

Erection of, illustrated by a Perspective View, Plans, Elevations, and 
Sections of a Pair of Villas, with the Specification, Quantities, and 
Estimates. By S. H. Brooks. 12mo, cloth 81.00 

FARM BUILDINGS: 

Their Arrangement and Construction, with Plans and Estimates. By 
Professor J. Scott. 12mo, cloth §q 

SHORING, 

And its Application. By G. K. Blagrove. Crown 8vo, cloth. t QQ 

ARCHES, PIERS, BUTTRESSES. 

By William Bland. 12mo, cloth t Q(y 

PRACTICAL BRICKLAYING. 

General Principles of Bricklaying: Arch Drawing. Cutting, and Setting: 
Pointing; Paving, Tiling, <kc. By Adam Hammond. With 68 "Wood- 
cuts. 12mo, cloth t QQ 

ART OF PRACTICAL BRICK=CUTTING AND SETTING. 

By Adam Hammond. With 90 Engravings. 12mo, cloth. . . . t QQ 

BRICKWORK: 

Embodying the General and Higher Principles of Bricklaying, Cutting, 
and Setting: with the Application of Geometry to Roof Tiling, <fec. 

By F. Walker. 12mo, cloth t QQ 

" Contains all that a student needs to learn from books. — Building News." 

BRICKS AND TILES, 

Rudimemarv Treatise on the Manufacture of. Containing an Outline 
of the PrinciDles of Brickmaking. By E. Dobson. M.R.I.B.A. Addi 
tions by C. Tomlinson, F.R.S. Illustrated. 12mo, cloth. . . .§1.20 

PRACTICAL BRICK AND TILE BOOK. 

Comprising: Brick and Tile Making, by E. Dobson, M.Inst.C.E. ; Prac- 
tical Bricklaying by A. Hammond, Brick-Cutting and Setting, by A. 
Hammond. 550 pp", with 270 Illustrations, strongly half -bound . S2.-40 

PRACTICAL MASONRY. 

A Guide to the Art of Stone Cutting. Comprising the Construction, 
Setting-Out, and Working of Stairs. Circular Work, Arches. Xiches. 
Domes. Pendentives. Vaults. Tracery Windows, &c: to which are 
added Supplements relating to Masonry Estimating and Quantity Sur- 
veying, and to Building Stones and Marbles, and a Glossary of Terms. 
Tor the Use of Students, Masons, and Craftsmen. By W. R. PrRCHASE, 
Building Inspector to the Borough of Hove. Fifth Edition, Enlarged. 
Royal 8vo, 226 pp., with 52 Plates, comprising over 400 Diagrams, 
cloth S3.00 

MASONRY AND STONECUTTING, 

The Principles of Masonic Projection, and their Application to Con- 
struction. By E. Dobson, M.R.I.B.A. 12mo, cloth SI. 00 

MODERN LIGHTNING CONDUCTORS. 

An Illustrated Suoplement to the Report of the Lightning Research 
Committee of 1905.' with Notes as to the Methods of Protection, and 
Specific at ions. Bv Klllingworth Hedges. M.Inst.C.E.. M.I.E.E., 
Honorary Secretary to the Lightning Research Committee, Author of 
" \merican Street Railwavs." Medium 8vo. cloth 

[J ust Published -^ S3.00 

"The illustrations are very interesting and give one a clear idea of what 

is likelv to happen when a building is struck by lightning. Mr. Hedges' 

suggestions of possible reasons why certain protected buildings were struck 



ARCHITECTURE, BUILDING, &*c. 33 

are instructive He also explains the modern methods of fitting buildings 
with lightning conductors. To the ordinary reader the book will be of in- 
terest, and to anyone who has to design a system for protecting a building 
from lightning strokes it will be helpful." — Builder. 

PLUMBING: 

A Text-Book to the Practice of the Art or Craft of the Plumber. With 
Chapters upon House Drainage and Ventilation. By Wm. Paton 
Buchan. Ninth Edition, with 512 Illustrations. Crown 8vo, cloth. 

$1.40 
HEATING BY HOT WATER, 

VENTILATION AND HOT WATER SUPPLY. 

By Walter Jones, M.I.M.E. 360 pages, with 140 Illustrations. 
Medium 8vo, cloth $2.50 

THE PRACTICAL PLASTERER: 

A Compendium of Plain and Ornamental Plaster Work. By W. Kemp. 
12mo, cloth .80 

CONCRETE: ITS NATURE AND USES. 

A Book for Architects, Builders, Contractors, and Clerks of Works. By 
G L Sutcliffe, A.R.I.B.A. Second Edition, Revised and Enlarged. 
396 pp., with Illustrations. 12mo, cloth. . . .[Just Published. $3,50 

PORTLAND CEMENT FOR USERS. 

By the late Henry Faija, M.Inst.C.E. Fifth Edition. Revised and 
Enlarged by D. B. Butler, A.M.Inst.C.E. 12mo, cloth $1.20 

LIMES, CEMENTS, MORTARS, CONCRETES, MASTICS, 

PLASTEPING &c. 

By G. R. Burnell, C.E. Fifteenth Edition. 12mo, eloth ,60 

MEASURING AND VALUING ARTIFICERS' WORK 

(The Student's Guide to the Practice of). Containing Directions for 
taking Dimensions, Abstracting the same, and bringing the Quantities 
into Bill, with Tables of Constants for Valuation of Labour, and for the 
Calculation of Areas and Solidities. Originally edited by E. Dobson, 
Architect. With Additions by E. W. Tarn, M.A. Seventh Edition, 
Revised. 12mo, cloth $3.00 

QUANTITIES AND MEASUREMENTS, 

In Bricklayers', Masons', Plasterers', Plumbers', Painters', Paper 
hangers'. Gilders', Smiths', Carpenters' and Joiners' Work. By A. C- 

Beaton, Surveyor. 12mo, cloth ,60 

"This book is indispensable to builders and their quantity clerks." — Eng* 
lish Mechanic. 

TECHNICAL GUIDE, MEASURER, AND ESTIMATOR. 

For Builders and Surveyors. Containing Technical Directions for Meas- 
uring Work in all the Building Trades, Complete Specifications for 
Houses, Roads, and Drains, and an Easy Method of Estimating the 
parts of a Building collectively. By A. C, Beaton. Tenth Edition. 

Waistcoat-pocket size ,60 

"No builder, architect, surveyor, or valuer should be without his 
'Beaton.' " — Building News. 

COMPLETE MEASURER; 

Setting forth the Measurement of Boards, Glass, Timber, and Stone. 
By R. Horton. Sixth Edition. 12mo, cloth $1.60 

ARCHITECTURAL PERSPECTIVE. 

The whole Course and Operations of the Draughtsman in Drawing a 
Large House in Linear Perspective. Illustrated by 43 Folding Plates. 
By F. O. Ferguson. Third Edition. 8vo, boards $1.50 



34 CROSBY LOCKWOOD er= SOX'S CATALOGUE. 
PERSPECTIVE FOR BEGINNERS 

For Students and Amateurs in Architecture, Painting, &e. By G. 
Pyne. Crown 8vo, cloth ,§() 

PRACTICAL RULES ON DRAWING. 

For the Builder and Young Student in Architecture. By G. Ptne. 4to 

S3.00 
THE MECHANICS OF ARCHITECTURE. 

A Treatise on Applied Mechanics, especially Adapted to the Use of 
Architects. By E. W. Tarn. M.A., Author of "The Science of Build- 
ing,'' &c. Second Edition, Enlarged. Illustrated with 125 Diagrams. 
12mo, cloth S3.00 

* ' The book is a very useful and helpful manual of architectural mechan- 
ics." — Builder. 

A HANDY BOOK OF VILLA ARCHITECTURE. 

Being a Series of Designs for Villa Residences in various Styles. "With 
Outline Specifications and Estimates. By C. Wickes. Architect, Au- 
thor of "The Spires and Towers of England," <fcc. 61 Plates, 4to, half- 
morocco, gilt edges SI 2. 00 

DECORATIVE PART OF CIVIL ARCHITECTURE. 

By Sir William Chambers, F.R.S. With Portrait, Illustrations, Notes, 
and an Examination of Grecian Architecture, by Joseph Gwilt, 
F.S.A. Revised and Edited bv W. H. Leeds. 66 Plates, 4to, cloth. 

S8.40 

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SANITATION AND WATER SUPPLY. 35 

VITRUVIUS— THE ARCHITECTURE OF MARCUS 
V1TRUVIUS POLLIO. 

In Ten Books. Translated from the Latin by J. Gwilt. With 23 
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THE WATER SUPPLY OF TOWNS AND THE CON- 

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36 CROSBY LOCKWOOD &» SON'S CATALOGUE. 
THE PURIFICATION OF SEWAGE. 

Being a Brief Account of the Scientific Principles of Sewage Purifica- 
tion, and their Practical Application. By Sidney Barwise, V.D 
(Lond.), B.Sc, M.R.C.S., D.P.H. (Camb.), Fellow of the Sanitary In- 
stitute, Medical Officer of Health to the Derbyshire County Council. 
Second Edition, Revised and Enlarged, with an Appendix on the Analy- 
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Diagrams. Demy 8vo, cloth Net §3.50 

Summary of Contexts: — Sewage: Its Nature axd Composition. — 
The Chemistry of Sewage. — Varieties of Sewage and the Changes it 
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tion of Sewage. — Principles involved in the Oxidation of Sewage. — 
Artificial Processes of Purification. — Automatic Distributors and 
Special Filters. — Particulars of Sewerage and Sewage Disposal 
Schemes required bt Local Government Board. — Useful Data. — Ap- 
pendix: The Apparatus required for Sewage Analysis. — Standard 
Solutions used in the Method of Sewage Analysis. — Tables: Esti- 
mation of Ammonia. — Nitrogen as Nitrates. — Incubator Test, Oxygen 
Absorbed. — To Convert Grains per Gallon to Parts per 100,000. 

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VENTILATION: 

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CARPENTRY, TIMBER, ETC. 



PRACTICAL FORESTRY. 

And its Bearing on the Improvement of Estates. By Charles E. 
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Estate Management, at the College of Agriculture, Downton. Second 

Edition, Revised. 12mo, cloth SI. 40 

Prefatory Remarks. — Objects of Planting. — Choice of a Forester. 
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Distances and Distribution of Trees in Plantations. — Trees an» 
Ground Game. — Attention after Planting. — Thinning of Plantations. 
— Pruning of Forest Trees. — Realization. — Methods of Sale. — 
Measurement of Timber. — Measurement and Valuation of Larch's 
Plantation. — Fire Lines. — Cost of Planting. 

WOODWORKING MACHINERY. 

Its Rise, Progress, and Construction. With Hints on the Management 
of Saw Mills and the Economical Conversion of Timber. Illustrated 
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CARPENTRY, TIMBER, &c. 37 

THE ELEMENTARY PRINCIPLES OF CARPENTRY. 

A Treatise on the Pressure and Equilibrium of Timber Framing, the 
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ties of Materials, &c. By Thomas Tredgold, C.E. With an Appendix 
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THE CARPENTER'S GUIDE. 

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THE JOINTS MADE AND USED BY BUILDERS. 

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38 CROSBY LOCKWOOD & SON'S CATALOGUE. 
THE PRACTICAL TIMBER MERCHANT. 

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PACKING-CASE TABLES. 

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GUIDE TO SUPERFICIAL MEASUREMENT. 

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DECORATIVE ARTS, ETC. 



SCHOOL OF PAINTING FOR THE IMITATION OF 
WOODS AND MARBLES. 

As Taught and Practised by A. R. Van der Burg and P. Van der 
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Net $10.00 
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Grains. — 11,12. Breche Marble; Preliminary Stages of Working and 
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Grains. — 14, 15. Bird's-Eye Maple; Preliminary Stages and Finished 
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Marble. — 17, 18. White Marble; Preliminary Stages of Process and 
Finished Specimen. — 19. Mahogany; Specimens of Various Grains and 
Methods of Manipulation. — 20. 21, Mahogany; Earlier Stages and 
Finished Specimen. — 22, 23, 24. Sienna Marble; Varieties of Grain, 
Preliminary Stages and Finished Specimen. — 25, 26, 27. Juniper Wood; 
Methods of Producing Grain, &c. ; Preliminary Stages and Finished 
Specimen. — 28, 29, 30. Vert de Mer Marble; Varieties of Grain and 
Methods of Working, Unfinished and Finished Specimens. — 31, 32, 33. 
Oak; Varieties of Grain, Tools Employed and Methods of Manipu- 
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ELEMENTARY DECORATION: 

As Applied to Dwelling-Houses, &c. By J. W. Facey. 12mo, cloth. 

80 

PRACTICAL HOUSE DECORATION. 

A Guide to the Art of Ornamental Painting, the arrangement of Colours 
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DECORATIVE ARTS, &c, 39 

ORNAMENTAL ALPHABETS, ANCIENT & MEDIAEVAL. 

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MODERN ALPHABETS, PLAIN AND ORNAMENTAL. 

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MEDIAEVAL ALPHABETS AND INITIALS. 

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A PRIMER OF THE ART OF ILLUMINATION. 

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THE EMBROIDERER'S BOOK OF DESIGN. 

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MARBLE DECORATION 

And the Terminology of British and Foreign Marbles. A Handbook 
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THE DECORATOR'S ASSISTANT. 

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GRAMMAR OF COLOURING. 

Applied to Decorative Painting and the Arts. By G. Field. New 
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HOUSE PAINTING, GRAINING, MARBLING, AND 

SIGN WRITING. 

With a Course of Elementary Drawing, and a Collection of Useful 
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12mo, cloth $2.00 

ART OF LETTER PAINTING MADE EASY. 

By J. G. Badenoch. With 12 full-page Engravings of Examples. 12mo, 

.60 



4© CROSBY LOCKWOOD &> SON'S CATALOGUE. 
PAINTING POPULARLY EXPLAINED. 

By Thomas John Gullick, Painter, and John Timbs, F.S.A. Includ- 
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GLASS STAINING, AND PAINTING ON GLASS. 

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WOOD=CARVING FOR AMATEURS. 

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NATURAL SCIENCE, ETC. 



THE VISIBLE UNIVERSE. 

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AN ASTRONOMICAL GLOSSARY. , 

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ASTRONOMY. 

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THE MICROSCOPE. 

Its Construction and Management. Including Technique, Photo-micro- 
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cloth S7.00 

MANUAL OF THE MOLLUSCA: 

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CHEMICAL MANUFACTURES, CHEMISTRY, &>c. 41 

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CHEMICAL MANUFACTURES, 
CHEMISTRY, ETC. 



THE OIL FIELDS OF RUSSIA AND THE RUSSIAN 

PETROLEUM INDUSTRY. 

A Practical Handbook on the Exploration, Exploitation, and Manage- 
ment of Russian Oil Properties, including Notes on the Origin of 
Petroleum in Russia, a Description of the Theory and Practice of 
Liquid Fuel, and a Translation of the Rules and Regulations concern- 
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THE ANALYSIS OF OILS AND ALLIED SUBSTANCES. 

By A. C.Wright, M.A.Oxon., B.Sc, Lond., formerly Assistant Lecturer 
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try at the Hull Technical School. Demy 8vo, cloth $3.50 

A HANDYBOOK FOR BREWERS. 

Being a Practical Guide to the Art of Brewing and Malting. Embracing 
the Conclusions of Modern Research which bear upon the Practice of 
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Enlarged. 530 pp. 12mo, cloth In press 

A POCKET=BOOK OF MENSURATION AND GAUGING. 

Containing Tables, Rules, and Memoranda for Revenue Officers, 
Brewers, Spirit Merchants, &c. By J. B. Mant, Inland Revenue. 
Second Edition, Revised. 18mo, leather $1.60 



42 CROSBY LOCKWOOD & SOX'S CATALOGUE. 
THE GAS ENGINEER'S POCKET=BOOK. 

Comprising Tables. Notes and Memoranda relating to the Manufacture, 
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By H. O'Coxxor. A.M. Inst.C.E. Second Edition, Revised. 470 pp., 
12mo. fully Illustrated, leather S3. 50 

LIGHTING BY ACETYLENE 

Generators, Burners, and Electric Furnaces. By William E. Gibbs, 
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ENGINEERING CHEMISTRY. 

A Practical Treatise for the Use of Analytical Chemists, Engineers, 
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NITRO=EXPLOSIYES. 

A Practical Treatise concerning the Properties, Manufacture, and 
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A HANDBOOK OF MODERN EXPLOSIYES. 

A Practical Treatise on the Manufacture and Use of Dynamite, Gun- 
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Their Sources and Properties, Modes of Storage and Transport. With 
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By H. Joshta Phillips, F.I.C., F.C.S. 12mo, 374 pp.. cloth, S3. 50 

"Merits a wide circulation, and an intelligent, appreciative study."— 
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A MANUAL OF THE ALKALI TRADE. 

Including the Manufacture of Sulphuric Acid, Sulphate of Soda, and 
Bleaching Powder. By Johx Lomas. Alkali Manufacturer. With 
232 Illustrations and Working Drawings. Second Edition, with 
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THE BLOW PIPE IN CHEMISTRY, MINERALOGY, Etc. 

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THE MANUAL OF COLOURS AND DYE=\YARES. 

Their Properties. Applications, Valuations, Impurities and Sophistica- 
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J. W. Slater. Second Edition, Revised and greatlv Enlarged. 12mo, 

cloth S3.00 

" There is no other work which covers precisely the same ground. To 
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INDUSTRIAL AND USEFUL ARTS. 43 

THE ARTISTS* MANUAL OF PIGMENTS. 

Showing their Composition, Conditions of Permanency, Non-Per- 
manency, and Adulterations, &c, with Tests of Purity. By H. C. 
Standage. Third Edition. 12mo, cloth SI. 00 

" This work is indeed multum-in-parvo , and we can, with good conscience, 
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dealers, or users. : — Chemical Review. 



INDUSTRIAL ARTS, TRADES, AND 
MANUFACTURES. 



THE CULTIVATION AND PREPARATION OF PARA 

RUBBER. 

By W. H. Johnson, F.L.S., F.R.H.S., Director of Agriculture, Gold 
Coast Colony, West Africa, Commissioned by Government in 1902 to 
visit Ceylon to Study the Methods employed there in the Cultivation 
and Preparation of Para Rubber and other Agricultural Staples for 
Market, with a view to Introduce them into West Africa. Demy 8vo, 

doth $3.00 

Summary op Contents: — Introductory. — The Para Rubber Tree 
(Hevea brasiliensis) at Home and Abroad. — Cultivation of the Tree: — 
Propagation. — Site for Plantation. — Distance Apart to Plant 
the Trees. — Transplanting. — Cultivation. — Insect Pests and 
Fungoid Diseases. — Collecting the Rubber: — Various Methods 
Employed in Tapping Rubber Trees. — Flow of Latex Increased by 
Wounding the Tree. — How to Tap. — The Preparation of Rubber 
from the Latex: — Latex. — Various Methods Employed in the Prep- 
aration of Rubber. — Suggested Method for Preparing Rubber. — 
Scrap Rubber. — Yield of Para Rubber from Cultivated Trees: — 
Ceylon. — Malay Peninsula. — Gold Coast, West Africa. — Establish- 
ment and Maintenance of a Para Rubber Plantation: — Ceylon. — 
Malay Peninsula. — Commercial Value of the Oil in Hevea Seeds. 

RUBBER HAND STAMPS 

And the Manipulation of Rubber. A Practical Treatise on the Manu- 
facture of India-rubber Hand Stamps, Small Articles of India-rubber, 
The Hektograph, Special Inks, Cements, and Allied Subjects. By 
T. O'Conor Sloane, A.M., Ph.D. With numerous llllustrations. 
Square 8vo, cloth $1.00 

PRACTICAL PAPER=/UAKING. 

A Manual for Paper-Makers and Owners and Managers of Paper-Mills. 
With Tables, Calculations, &c. By G. Clapperton, Paper-Maker. 
With Illustrations of Fibres from Micro-Photographs. 12mo, cloth, 

$2.50 
THE ART OF PAPER=MAKING. 

A Practical Handbook of the Manufacture of Paper from Rags, Esparto, 
Straw, and other Fibrous Materials. Including the Manufacture of 
Pulp from Wood Fibre, with a Description of the Machinery and 
Appliances used. To which are added Details of Processes for Recover- 
ing Soda from Waste Liquors. By Alexander Watt. With Illus- 
trations. 12mo, cloth $3.00 

A TREATISE ON PAPER. 

For Printers and Stationers. With an Outline of Paper Manufacture; 
Complete Tables of Sizes, and Specimens of Different Kinds of Paper. 
By Richard Parkinson, late of the Manchester Technical School. 
Demy 8vo, cloth $1,40 



44 CROSBY LOCKWOOD &» SON'S CATALOGUE. 
THE ART OF SOAP=MAKING. 

A Practical Handbook of the Manufacture of Hard and Soft Soaps, 
Toilet Soaps, &c. Including many new Processes, and a Chapter on 
the Recovery of Glycerine from Waste Leys. By Alexander Watt. 
Sixth Edition, including an Appendix on Modern Candlemaking. 

12mo, cloth $3.00 

"A thoroughly practical treatise. We congratulate the author on the 

success of his endeavour to fill a void in English technical literature." — 

Nature, j 

"The work will prove very useful, not merely to the technological student, 
but to the practical soap boiler who wishes to understand the theory of his 
art." — Chemical News. 

THE ART OF LEATHER MANUFACTURE. 

A Practical Handbook, in which the Operations of Tanning, Currying, 
and Leather Dressing are fully Described, and the Principles of Tanning 
Explained. Together with a Description of the Arts of Glue Boiling, 
Gut Dressing, &c. By Alexander Watt. Fifth Edition, thoroughly 
Revised and Enlarged. 8vo, cloth Nearly ready, $4.00 

ART OF BOOT AND SHOEMAKING, 

Including Measurement, Last-fitting, Cutting-out, Closing, and Making; 
with a Description of the most Approved Machinery employed. By 
J. B. Leno. 12mo, cloth # gQ 

"By far the best work ever written on the subject." — Scottish Leather 
Trader. 

COTTON MANUFACTURE. 

A Manual of Practical Instruction of the Processes of Opening, Carding, 
Combing, Drawing, Doubling and Spinning of Cotton, the Methods of 
Dyeing, &c. For the use of Operatives, Overlookers, and Manu- 
facturers. By John Lister, Technical Instructor, Pendleton. 8vo, 
cloth, $3.00 

"A distinct advance in the literature of cotton manufacture." — Machinery 
"It is thoroughly reliable, fulfilling nearly all the requirements desired." 
Glasgow Herald. 

WATCH REPAIRING, CLEANING, AND ADJUSTING. 

A Practical Handbook dealing with the Materials and Tools Used and 
the Methods of Repairing, Cleaning, Altering, and Adjusting all kinds 
of English and Foreign Watches, Repeaters, Chronographs, and Marine 
Chronometers. By F. J. Garrard, Springer and Adjuster of Marine 
Chronometers and Deck Watches for the Admiralty. With over 200 
Illustrations. 12mo, cloth $2 t QQ 

MODERN HOROLOGY, IN THEORY AND PRACTICE. 

Translated from the French of Claudius Saunier, ex-Director of the 
School of Horology at Macon, by Julien Tripplin, F.R.A.S., Besancon 
Watch Manufacturer, and Edward Rigg, M.A., Assayer in the Royal 
Mint. With Seventy-eight Woodcuts and Twenty-two Coloured 
Copper Plates. Second Edition. Super-royal, 8vo, cloth . . . . $15.00 

Half -calf. $18.00 

THE WATCHMAKER'S HANDBOOK. 

Intended as a Workshop Companion for those engaged in Watchmaking 
and the Allied Mechanical Arts. Translated from the French of 
Claudius Saunier, and enlarged by Julien Tripplin, F.R.A.S., and 
Edward Rigg, M.A., Assayer in the Royal Mint. Fourth Edition 
12mo, clotb, .,,,,,, , . .$3.00 



INDUSTRIAL AND USEFUL ARTS. 45 

CLOCKS, WATCHES, & BELLS for PUBLIC PURPOSES. 

A Rudimentary Treatise. By Edmund Beckett, Lord Grimthorpe, 
LL.D., K.C, F.Jt.A.S. Eighth Edition, with new List of Great Bells 
and an Appendix on Weathercocks. 12mo, cloth $1.80 

HISTORY OF WATCHES & OTHER TIMEKEEPERS. 

By James F. Kendal, M.B.H.Inst. ,Q{) boards; or cloth, gilt, $1.00 

ELECTROPLATING & ELECTRO=REFINING of METALS, 

Being a new edition of Alexander Watt's "Electro-Deposition." 
Revised and Largely Rewritten by Arnold Philip, B.Sc, A.I.E.E., 
Principal Assistant to the Admiralty Chemist. 8vo, cloth. . . .$4.50 

ELECTROPLATING. 

A Practical Handbook on the Deposition of Copper, Silver, Nickel, Gold, 
Aluminium, Brass, Platinum, &c, &c. By J. W. Urquhart, C.E. 
Fifth Edition, Revised. 12mo, cloth $2.00 

ELECTRO=METALLURGY, 

Practically Treated. By Alexander Watt. Tenth Edition, enlarged 
and revised. With Additional Illustrations, and including the most 
Recent Processes. 12mo, cloth $1.40 

GOLDSMITH'S HANDBOOK, 

Containing full Instructions in the Art of Alloying, Melting, Reducing, 
Colouring, Collecting, and Refining. The Processes of Manipulation, 
Recovery of Waste, Chemical and Physical Properties of Gold; Solders, 
Enamels, and other useful Rules and Recipes, &c. By George E. 
Gee. Sixth Edition. 12mo, cloth $1.20 

SILVERSMITH'S HANDBOOK, 

On the same plan as the above. By George E. Gee. Third Edition. 
12mo, cloth $1.20 

*** The two preceding Works, in One handsome Volume, half-bound, en- 
titled "The Goldsmith's and Silversmith's Complete Handbook," $2.80 

JEWELLER'S ASSISTANT IN WORKING IN GOLD. 

A Practical Treatise for Masters and Workmen, Compiled from the 
Experience of Thirty Years' Workshop Practice. By George E. Gee. 
12mo $3.00 

HALL=MARKING OF JEWELLERY. 

Comprising an account of all the different Assay Towns of the United 
Kingdom, with the Stamps at present employed ; also the Laws relating 
to the Standards and Hall-marks at the Various Assay Offices. By 
George E. Gee. 12mo, cloth $1.20 

ELECTROTYPING. 

The Reproduction and Multiplication of Printing Surfa»es and Works 
of Art by the Electro-Deposition of Metals. By J. W. Urquhart, C.E. 
12mo, cloth $2.00 

MECHANICAL DENTISTRY: 

A Practical Treatise on the Construction of the Various Kinds of 
Artificial Dentures, comprising also Useful Formula?, Tables and 
Receipts. By C. Hunter. 12mo, cloth $1.20 

BRASS FOUNDER'S MANUAL: 

Instructions for Modelling, Pattern Making, Moulding, Turning, &c. 
By W. Graham. 12mo, cloth # §q 



46 CROSBY LOCKWOOD & SON'S CATALOGUE 
SHEET METAL WORKER'S INSTRUCTOR. 

Comprising a Selection of Geometrical Problems and Practical Rules 
for Describing the Various Patterns Required by Zinc. Sheet-Iron, 
Copper, and Tin-Plate Workers. By Reuben Henry Warn. Piactical 
Tin-Plate Worker. New Edition, Revised and greatly Enlarged by 
Joseph G. Horner, A.M.I.M.E. 12mo, 254 pp.. with 430 Illustra- 
tions, cloth §3.00 

SHEET METAL=WORKER'S GUIDE. 

A Practical Handbook for Tinsmiths, Coppersmiths, Zincworkers, &c , 
with 46 Diagrams and Working Patterns. By W. J. E. Crane. Fourth 
Edition. 12mo, cloth # gQ 

GAS FITTING: 

A Practical Handbook. By John Black. Revised Edition With 
130 Illustrations. 12mo, cloth $1.00 

" It is written in a simple, practical style, and we heartily recommend it." 
— Plumber and Decorator. 

TEA MACHINERY AND TEA FACTORIES. 

A Descriptive Treatise on the Mechanical Appliances required in the 
Cultivation of the Tea Plant and the Preparation of Tea for the Market. 
By A. J. Wallis-Tayler, A.M.Inst C.E. Medium 8vo, 468 pp. With 
218 Illustrations .$10.00 

Summary of Contents. 

Mechanical Cultivation or Tillage of the Soil. — Plucking or 
Gathering the Leaf. — Tea Factories. — The Dressing, Manufacture, 
or Preparation of Tea by Mechanical Means. — Artificial Wither- 
ing of the Leaf. — Machines for Rolling or Curling the Leaf. — Fer- 
menting Process. — Machines for the Automatic Drying or Firing 
of the leaf. — Machines for Non-Automatic Drying or Firing of the 
Leaf. — Drying or Firing Machines. — Breaking or Cutting, and Sort- 
ing Machines. — Packing the Tea. — Means of Transport on Tea Plan- 
tations. — Miscellaneous Machinery and Apparatus. — Final Treat- 
ment of the Tea. — Tables and Memoranda. 

FLOUR MANUFACTURE. 

A Treatise on Milling Science and Practice. By Friedrich Kick, 
Imperial Regierungsrath, Professor of Mechanical Technology in the 
Imperial German Polytechnic Institute, Prague. Translated from the 
Second Enlarged and Revised Edition. By H. H. P. Powles, A.M. 
Inst. C.E. 400 pp., with 28 Folding Plates, and 167 Woodcuts. Royal 
8vo, cloth $10.00 

ORNAMENTAL CONFECTIONERY. 

A Guide for Bakers, Confectioners and Pastrycooks; including a 
variety of Modern Recipes, and Remarks on Decorative and Coloured 
Work. With 129 Original Designs. By Robert Wells. 12mo, cloth, 

S3. 00 
BREAD & BISCUIT BAKER'S & SUGAR=BOILER'S 

ASSISTANT. 

Including a large variety of Modern Recipes. With Remarks on the 
Art of Bread-making. By Robert Wells. Fourth Edition. 12mo, 
cloth .50 

PASTRYCOOK & CONFECTIONER'S GUIDE. 

For Hotels, Restaurants, and the Trade in general, adapted also for 
Family Use. By R. Wells, Author of "The Bread and Biscuit Baker." 

.40 



INDUSTRIAL AND USEFUL ARTS. 47 

MODERN FLOUR CONFECTIONER. 

Containing a large Collection of Recipes for Cheap Cakes, Biscuits, &c. 
With remarks on the Ingredients Used in their Manufacture. By R. 
Wells 40 

SAVOURIES AND SWEETS 

Suitable for Luncheons and Dinners. By Miss M. L. Allen (Mrs. A. 
Macaire), Author of "Breakfast Dishes," &c. Thirtieth Edition. F'cap 
8vo, sewed .40 

BREAKFAST DISHES 

For Every Morning of Three Months. By Miss Allen (Mrs. A. 
Macaire), Author of "Savouries and Sweets," &c. Twenty-second 
Edition. F'cap 8vo, sewed .40 

MOTOR CARS OR POWER CARRIAGES FOR COMMON 
ROADS. 

By A. J. Wallis-Tayler, A.M.Inst.C.E. 12mo, cloth $1.80 

FRENCH POLISHING AND ENAMELLING. 

A Practical Book of Instruction, including numerous Recipes from mak- 
ing Polishes, Varnishes, Glaze Lacquers, Revivers, &c. By R. Bit- 
mead. 12mo, cloth .00 

CEMENTS, PASTES, GLUES, AND GUMS. 

A Guide to the Manufacture and Application of Agglutinants for 
Workshop, Laboratory, or Office Use. With 900 Recipes and Formulae. 
By H. C. Standage. Crown 8vo, cloth ,80 

PRACTICAL ORGAN BUILDING. 

By W. E. Dickson, M.A., Precentor of Ely Cathedral. Second Edition, 
Revised. 12mo, cloth $1.00 

COACH=BUILDING: 

A Practical Treatise, Historical and Descriptive. By. J. W. Burgess. 
12mo, cloth $1.00 

SEWING MACHINERY. 

Construction, History, Adjusting, &c. By J. W. Urquhart. 12mo, 
cloth .80 

WOOD ENGRAVING: 

A Practical and Easy Introduction to the Study of the Art. By W. N. 
Brown. 12mo, cloth .00 

LAUNDRY MANAGEMENT. 

A Handbook for Use in Private and Public Laundries. 12mo, cloth 

.80 
CONSTRUCTION OF DOOR LOCKS. 

From the Papers of A. C. Hobbs. Edited by Charles Tomlinson, 
F.R.S. With a Note upon Iron Safes by Robert Mallet. 12mo, 
cloth $1.00 



48 CROSBY LOCKWOOD & SON'S CATALOGUE. 

HANDYBOOKS FOR HANDICRAFTS. 

BY PAUL N. HASLUCK, 

Author of "Lathe Work," &c. 12mo, 144 pp., price 50c. each. 
B3F 2 * These Handybooks havebeen written to supply information for Work- 
men, Students, and Amateurs in the several Handicrafts, on the actual 
Practice of the Workshop, and are intended to convey in plain language 
Technical Knowledge of the several Crafts. In describing the processes 
employed , and the manipulation of material, workshop terms are used; work- 
shop practice is fully explained; and the text is freely illustrated with drawings 
of modem tools, appliances, and processes. 



METAL TURNER'S HANDYBOOK. 

A Practical Manual for Workers at the Foot-Lathe. With 100 Illus- 
trations .50 

WOOD TURNER'S HANDYBOOK. 

A Practical Manual for Workers at the Lathe. With over 100 Illus- 
trations .50 

WATCH JOBBERS HANDYBOOK. 

A Practical Manual on Cleaning, Repairing, and Adjusting. With 
upwards of 100 Illustrations ,50 

PATTERN MAKER'S HANDYBOOK. 

A Practical Manual on the Construction of Patterns for Founders. 
With upwards of 100 Illustrations .50 

MECHANIC'S WORKSHOP HANDYBOOK. 

A Practical Manual on Mechanical Manipulation, embracing Informa- 
tion on various Handicraft Processes. With Useful Notes and Mis- 
cellaneous Memoranda. Comprising about 200 Subjects ,50 

MODEL ENGINEER'S HANDYBOOK. 

A Practical Manual on the Construction of Model Steam Engines. 
With upwards of 100 Illustrations .50 

CLOCK JOBBER'S HANDYBOOK. 

A Practical Manual on Cleaning, Repairing, and Adjusting. With 
upwards of 100 Illustrations .50 

CABINET WORKER'S HANDYBOOK. 

A Practical Manual on the Tools, Materials, Appliances, and Processes 
employed in Cabinet Work. With upwards of 100 Illustrations. .50 

*'Mr. Hasluck's thorough-going little Handybook is amongst the most 
practical guides we have seen for beginners in cabinet-work." — Saturday 
Review. 

WOODWORKER'S HANDYBOOK. 

Embracing Information on the Tools, Materials, Appliances, and 
Processes Employed in Woodworking. With 104 Illustrations. .50 



COMMERCE, COUNTING-HOUSE WORK, &c. 49 



COMMERCE, COUNTING-HOUSE WORK, 
TABLES, ETC. 



LESSONS IN COMMERCE. 

By Professor R. Gambaro, of the Royal High Commercial School at 
Genoa. Edited and Revised by James Gault, Professor of Commerce 
and Commercial Law in King's College, London. Fifth Edition. 
12mo, cloth $1.40 



THE FOREIGN COMMERCIAL CORRESPONDENT. 

Being Aids to Commercial Correspondence in Five Languages — English, 
French, German, Italian, and Spanish. By Conrad E. Baker. Third 
Edition, Carefully Revised Throughout. 12mo, cloth $1.80 



FACTORY ACCOUNTS: their PRINCIPLES & PRACTICE. 

A Handbook for Accountants and Manufacturers, with Appendices on 
the Nomenclature of Machine Details; the Income Tax Acts; the 
Rating of Factories ; Fire and Boiler Insurance ; the Factory and Work- 
shop Acts, &c, including also a Glossary of Terms and a large numbei 
of Specimen Rulings. By Emile Garcke and J. M. Fells. Fifth 
Edition, Revised and Enlarged. Demy 8vo, cloth $3.00 



MODERN METROLOGY. 

A Manual of the Metrical Units and Systems of the present Century. 
With an Appendix containing a proposed English System. By 
Lowis d'A. Jackson, A.M. Inst. C.E., Author of "Aid to Survey Prac- 
tice," &c. 8vo, cloth $5.00 



SERIES OF METRIC TABLES. 

In which the British Standard Measures and Weights are compared 
with those of the Metric System at present in Use on the Continent. 
By C. H. Dowling, C.E. 8vo, cloth $4.00 



IRON-PLATE WEIGHT TABLES. 

For Iron Shipbuilders, Engineers, and Iron Merchants Containing the 
Calculated Weights of upwards of 150,000 different sizes of Iron Plates 
from 1 foot by 6 in. by ± in. to 10 feet by 5 feet by 1 in. Worked out 
on the Basis of 40 lbs. to the square foot of Iron of 1 inch in thickness. 
By H. Burlinson and W. H. Simpson. 4to, half -bound. . . .$10.00 



50 CROSBY LOCKWOOD 6- SON'S CATALOGUE. 

AGRICULTURE, FARMING, 
GARDENING, ETC. 



THE COMPLETE GRAZIER AND FARMER'S AND 

CATTLE BREEDER'S ASSISTANT. 

A Compendium of Husbandry. Originally Written by William 
Youatt. Fourteenth Edition, entirely Re-written, considerably En- 
larged, and brought up to Present Requirements, by William Fream, 
LL.D., Assistant Commissioner, Royal Commission on Agriculture, 
Author of "The Elements of Agriculture," &c. Royal, 8vo, 1,100 pp., 
450 Illustrations, handsomely bound $12.00 

STOCK: CATTLE, SHEEP, AND HORSES. 

Vol. III.— OUTLINES OF MODERN FARMING. By R. Scott Burn. 
Woodcuts. 12mo, cloth $1.00 

SHEEP: 

The History, Structure, Economy, and Diseases of. By W. C. Spooner. 
Fifth Edition, with Engravings, including Specimens of New and 
Improved Breeds. 12mo, cloth $1.40 

MEAT PRODUCTION: 

/ Manual for Producers, Distributors, and Consumers of Butchers' 
i^eat. By John Ewart. 12mo, cloth $1.00 

MILK, CHEESE, AND BUTTER. 

A Practical Handbook on their Properties and the Processes of their 
Production. Including a Chapter on Cream and the Methods of its 
Separation from Milk. By John Oliver, late Principal of the Western 
Dairy Institute, Berkeley. With Coloured Plates and 200 Illustra- 
tions. 12mo, cloth $3.00 

BRITISH DAIRYING. 

A Handy Volume on the Work of the Dairy-Farm. For the Use of 
Technical Instruction Classes, Students in Agricultural Colleges and 
the Working Dairy-Farmer. By Prof. J. P. Sheldon. With Illus- 
trations. Second Edition, Revised. 12mo, cloth $1.00 

DAIRY, PIGS, AND POULTRY. 

Vol. IV. OUTLINES OF MODERN FARMING. By R. Scott 
Burn. Woodcuts. 12mo, cloth .80 

THE ELEMENTS OF AGRICULTURAL GEOLOGY. 

A Scientific Aid to Practical Farming. By Primrose McConnell. 
Author of "Note-book of Agricultural Facts and Figures." 8vo, cloth, 

$7.50 
SOILS, MANURES, AND CROPS. 

Vol. I.— OUTLINES OF MODERN FARMING. By R. Scott Burn. 
Woodcuts. 12mo, cloth .80 

FERTILISERS AND FEEDING STUFFS. 

Their Properties and Uses. A Handbook for the Practical Farmer. 
By Bernard Dyer, D.Sc. (Lond.) With the Text of the Fertilisers 
and Feeding Stuffs Act of 1893, The Regulations and Forms of the 
Board of Agriculture, and Notes on the Act by A. J. David, B.A., 
LL.M. Fourth Edition, Revised. 12mo, cloth .40 

THE ROTHAMSTED EXPERIMENTS AND THEIR 

PRACTICAL LESSONS FOR FARMERS. 

Part I. Stock. Part II. Crops. By C. J. R. Tipper. 12mo, cloth 

$1.40 



AGRICULTURE, FARMING, GARDENING, &c. 51 
SYSTEMATIC SMALL FARMING. 

Or, The Lessons of My Farm. Being an Introduction to Modern Farm 
Practice for Small Farmers. By R. Scott Burn, Author of "Outlines 
of Modern Farming," &c. 12mo, cloth $2.40 

THE FIELDS OF GREAT BRITAIN. 

A Text-Book of Agriculture. Adapted to the Syllabus of the Science 
and Art Department. For Elementary and Advanced Students. By 
Hugh Clements (Board of Trade). Second Edition, Revised, "with 
Additions. 18mo, cloth $1.00 

OUTLINES OF MODERN FARMING. 

By R. Scott Burn. Soils, Manures, and Crops — Farming and Farming 
Economy — Cattle, Sheep, and Horses — Management of Dairy, Pigs, 
and Poultry — Utilisation of Town-Sewage, Irrigation, &c. Sixth 
Edition. In One Vol., 1,250 pp., half -bound, profusely Illustrated. 

$4.80 
FARM ENGINEERING, The COMPLETE TEXT=B00K of. 

Comprising Draining and Embanking; Irrigation and Water Supply; 
Farm Roads, Fences and Gates; Farm Buildings; Barn Implements 
and Machines ; Field Implements and Machines ; Agricultural Survey- 
ing, &c. By Professor John Scott. 1,150 pp., half-bound, with over 
600 Illustrations $4.80 

DRAINING AND EMBANKING. 

A Practical Treatise. By John Scott, late Professor of Agriculture 
and Rural Economy at the Royal Agricultural College, Cirencester. 
With 68 Illustrations. 12mo, cloth .60 

"A valuable handbook to the engineer as well as to the surveyor." — Land. 

IRRIGATION AND WATER SUPPLY: 

A Practical Treatise on Water Meadows, Sewage Irrigation, Warping, 
&c; on the Construction of Wells, Ponds, and Reservoirs, &c. By 
Professor J. Scott. 12mo, cloth 60 

FARM ROADS, FENCES, AND GATES: 

A Practical Treatise fon the Roads, Tramways, and Waterways of the 
Farm ; the Principles of Enclosures ; and on Fences, Gates, and Stiles. 
By Professor John Scott. 12mo, cloth .60 

BARN IMPLEMENTS AND MACHINES: 

Treating of the Application of Power to the Operations of Agriculture 
and of the various Machines used in the Threshing-barn, in the Stock- 
yard, Dairy, &c. By Professor John Scott. With 123 Illustrations. 
12mo, cloth .80 

FIELD IMPLEMENTS AND MACHINES: 

With Principles and Details of Construction and Points of Excellence, 
their Management, &c. By Professor John Scott. With 138 Illus- 
trations. 12mo, cloth .80 

AGRICULTURAL SURVEYING: 

A Treatise on Land Surveying, Levelling, and Setting-out* with Direc- 
tions for Valuing and Reporting on Farms and Estates. By Professor 
J. Scott. 12mo, cloth 60 

OUTLINES OF FARM MANAGEMENT. 

Treating of the General Work of the Farm; Stock* Contract Work, 
Labour, &c. By R. Scott Burn. 12mo, cloth $1.00 

OUTLINES OF LANDED ESTATES MANAGEMENT. 

Treating of the Varieties of Lands, Methods of Farming, the Setting-out 
of Farms, &c; Roads, Fences, Gates, Irrigation, Drainage, &c. By 
R. S. Burn. 12mo, cloth $1.00 



52 CROSBY LOCKAXOOD & SOX'S CATALOGUE. 
FARMING AND FARMING ECONOMY. 

Historical and Practical. Vol. II. — OUTLINES OF MODERN 
FARMING. By R. Scott Bcrx. 12mo, cloth SI. 20 

UTILIZATION OF SEWAGE, IRRIGATION, &c. 

Vol. V.— OUTLINES OF MODERN FARMING. By R. Scott Bourn. 
Woodcuts. 12mo, cloth 81.00 

N0TE=B00K OF AGRICULTURAL FACTS & FIGURES 

FOR FARMERS AND FARM STUDENTS. 

By Primrose McCoxxell, B.Sc, Fellow of the Highland and Agri- 

. cultural Society, Author of "Elements of Farming." Seventh Edition, 

Re-written, Revised, and greatly Enlarged. leap. 8vo, 480 pp., 

leather, gilt edges [Just Published. §3.00 

TABLES and .MEMORANDA for FARMERS, GRAZIERS, 
AGRICULTURAL STUDENTS, SURVEYORS, LAND AGENTS, 
AUCTIONEERS, &c. 

With a New System of Farm Book-keeping. By Sidxey Fraxcis. 
Fifth Edition. 272 pp., waistcoat-pocket size, limp leather. . . ,60 

THE HAY AND STRAW MEASURER: 

New Tables for the Use of Auctioneers, Valuers, Farmers, Hay and 
Straw Dealers, &c, forming a complete Calculator and Ready Reck- 
oner. By Johx Steele. 12mo, cloth # §Q 

READY RECKONER FOR ADMEASUREMENT OF LAND. 

By A. Armax. Revised and extended by C. Norms, Surveyor. Fifth 
Edition. 12mo, cloth ,§0 

THE HORTICULTURAL N0TE=B00K. 

A Manual of Practical Rules, Data, and Tables, for the use of Students, 
Gardeners, Nurserymen, and others interested in Flower, Fruit, and 
Vegetable Culture, or in the Laying-out and Management of Gardens. 
By J. C. NewsHAM, F.R..H.S., Headmaster of the Hampshire County 
Council Horticultural School. "With numerous Illustrations. Fcap. 
Svo, cloth [Just Published. S3. 00 

MARKET AND KITCHEN GARDENING. 

By C. W. Shaw, late Editor of "Gardening Illustrated." Crown Svo, 

81.40 
A PLAIN GUIDE TO GOOD GARDENING; 

Or, How to Grow Vegetables. Fruits, and Flowers. By S. Wood. 
Fourth Edition, with considerable Additions, and numerous Illustra- 
tions. 12mo, cloth SI. 40 

THE FORCING GARDEN; 

Or, How to Grow Early Fruits. Flowers and Vegetables. With Plans 
and Estimates for Building Glasshouses, Pits and Frames. With 
Illustrations. By Samcel"Wood. 12mo, cloth SI. 40 

KITCHEN GARDENING MADE EASY. 

Showing the best means of Cultivating even. 1 known Vegetable and 
Herb, &c, with directions for management all the year round. Bv 
Geo. M. F. Glexxt. Illustrated. 12mo, cloth .60 

COTTAGE GARDENING; 

Or, Flowers, Fruits, and Vegetables for Small Gardens. By E. Hobday. 
12mo, cloth .60 

GARDEN RECEIPTS. 

Edited by Charles W. Qctx. Fourth Edition. 12mo, cloth, ,60 

MULTUM=IN=PARYO GARDENING ; 

Or, How to Make One Acre of Land produce $3007 a year, by the 
Cultivation of Fruits and Vegetables: al=o. How to Grow Flowers in 
Three Glass Houses, so as to realise S853.60 per annum clear Profit. 
By Samttel Wood, Author of "Good Gardening," &c. Sixth Edition. 
12mo, paper .50 



